💡Okay, so I have this problem with two parts, both about proving conditions for three points being the vertices of an equilateral triangle. Let me try to tackle them one by one.Starting with part (a). It says: Let ε = 1/2 + i√3/2. Prove that the points a, b, and c are the vertices of an equilateral triangle if and only if a + ε²b + ε⁴c = 0 or a + ε⁴b + ε²c = 0.Hmm, ε is given as 1/2 + i√3/2. I remember that this is one of the cube roots of unity. Cube roots of unity satisfy ε³ = 1, and they are 1, ε, and ε². Also, I recall that 1 + ε + ε² = 0. So, ε² is the complex conjugate of ε, right? Because ε = e^(2πi/3) and ε² = e^(4πi/3), which is the conjugate.So, if a, b, c form an equilateral triangle, then the rotation by 120 degrees (which is multiplication by ε) should relate the points. Maybe I can express one point in terms of the others using ε.Let me think. If I rotate point b around point a by 120 degrees, I should get point c. In complex numbers, rotation is multiplication by ε. So, c - a = ε(b - a). Let me write that down:c - a = ε(b - a)Expanding this, c = a + ε(b - a) = εb + (1 - ε)a.Similarly, if I rotate in the other direction, it would be multiplication by ε². So, maybe c = ε²b + (1 - ε²)a.Wait, but the problem states a + ε²b + ε⁴c = 0 or a + ε⁴b + ε²c = 0. That seems a bit different. Maybe I need to manipulate my equation to match this form.From c = εb + (1 - ε)a, let's rearrange it:c - εb = (1 - ε)aSo, a = (c - εb)/(1 - ε)But 1 - ε is 1 - (1/2 + i√3/2) = 1/2 - i√3/2, which is ε². So, a = (c - εb)/ε²Multiplying both sides by ε²:ε²a = c - εbBring c to the other side:ε²a + εb = cHmm, not quite matching the given condition. Maybe I need to consider another rotation.Alternatively, let's consider all three points. If the triangle is equilateral, then the distances between each pair of points are equal. So, |a - b| = |b - c| = |c - a|.But how does that relate to the given condition? Maybe using complex numbers properties.I remember that for three points to form an equilateral triangle, the following relation holds: a + εb + ε²c = 0 or a + ε²b + εc = 0. Wait, that's similar to what's given here, except here it's ε² and ε⁴. But since ε³ = 1, ε⁴ = ε. So, ε⁴ is the same as ε.Wait, ε⁴ = ε^(3+1) = ε^3 * ε = 1 * ε = ε. So, ε⁴ = ε. Similarly, ε² is just ε².So, the given conditions are a + ε²b + εc = 0 or a + εb + ε²c = 0. Which is the same as the standard condition for equilateral triangles in complex plane.So, maybe the problem is just expressing the same condition with ε⁴ instead of ε, but since ε⁴ = ε, it's the same thing.So, to prove the equivalence, I need to show that if a, b, c are vertices of an equilateral triangle, then one of these equations holds, and conversely, if one of these equations holds, then the triangle is equilateral.Let me try the forward direction first. Assume a, b, c form an equilateral triangle. Then, without loss of generality, we can assume that c is obtained by rotating b around a by 120 degrees, which is multiplication by ε. So, c - a = ε(b - a), which gives c = a + ε(b - a) = εb + (1 - ε)a.Rearranging, we get ε²a + εb + c = 0? Wait, let me see.Wait, from c = εb + (1 - ε)a, let's bring all terms to one side:c - εb - (1 - ε)a = 0Which is c - εb - a + εa = 0Factor out ε from the last two terms:c - εb - a + εa = c - a - εb + εa = (c - a) + ε(a - b) = 0Hmm, not sure if that helps. Alternatively, let me write it as:c = εb + (1 - ε)aMultiply both sides by ε²:ε²c = ε³b + ε²(1 - ε)aBut ε³ = 1, so ε²c = b + ε²(1 - ε)aBring all terms to one side:ε²c - b - ε²(1 - ε)a = 0Factor out ε²:ε²(c - (1 - ε)a) - b = 0Not sure. Maybe another approach.Alternatively, consider that in an equilateral triangle, the centroid is also the circumcenter, so the sum of the vertices multiplied by certain weights equals zero.Wait, I think the condition a + εb + ε²c = 0 comes from the fact that the centroid is at zero when considering the rotation. Let me recall that.If we have three points a, b, c forming an equilateral triangle, then the relation a + εb + ε²c = 0 holds when the centroid is at the origin. But in general, if the centroid is not at the origin, we can translate the system so that the centroid is at the origin.Wait, maybe I need to consider the centroid. The centroid G is (a + b + c)/3. If we translate the system so that G is at the origin, then a + b + c = 0. But in our case, the condition is a + ε²b + ε⁴c = 0. Since ε⁴ = ε, it's a + ε²b + εc = 0.So, if a + ε²b + εc = 0, then the centroid is not necessarily at the origin, but the weighted sum is zero. Hmm.Alternatively, maybe consider the vectors from one vertex to the others. For an equilateral triangle, the vector from a to b rotated by 60 degrees should give the vector from a to c.So, (b - a) rotated by 60 degrees is (c - a). Rotation by 60 degrees is multiplication by e^(iπ/3) = 1/2 + i√3/2, which is ε. So, (c - a) = ε(b - a).So, c - a = ε(b - a)Then, c = a + ε(b - a) = εb + (1 - ε)aRearranging, c - εb = (1 - ε)aSo, a = (c - εb)/(1 - ε)But 1 - ε is 1 - (1/2 + i√3/2) = 1/2 - i√3/2 = ε².So, a = (c - εb)/ε²Multiply both sides by ε²:ε²a = c - εbBring all terms to one side:ε²a + εb - c = 0Which is ε²a + εb + (-1)c = 0But the problem states a + ε²b + ε⁴c = 0. Wait, ε⁴ is ε, so it's a + ε²b + εc = 0.Comparing with what I have: ε²a + εb - c = 0.Hmm, not exactly the same. Maybe I need to consider a different rotation.Alternatively, if I rotate in the other direction, which would be multiplication by ε².So, (c - a) = ε²(b - a)Then, c = a + ε²(b - a) = ε²b + (1 - ε²)aRearranging, c - ε²b = (1 - ε²)aSo, a = (c - ε²b)/(1 - ε²)But 1 - ε² is 1 - (1/2 - i√3/2) = 1/2 + i√3/2 = ε.So, a = (c - ε²b)/εMultiply both sides by ε:εa = c - ε²bBring all terms to one side:εa + ε²b - c = 0Which is εa + ε²b + (-1)c = 0Again, comparing with the given condition a + ε²b + εc = 0, it's similar but not the same.Wait, maybe I need to factor out ε or something.Let me see, from εa + ε²b - c = 0, multiply both sides by ε²:ε³a + ε⁴b - ε²c = 0But ε³ = 1, so:a + εb - ε²c = 0Hmm, that's a + εb - ε²c = 0, which is different from the given condition.Wait, maybe I need to consider both possibilities of rotation, clockwise and counterclockwise, which would give two different equations.So, if I rotate b around a by 60 degrees to get c, I get one equation, and if I rotate b around a by -60 degrees (which is 300 degrees), I get another equation.So, in the first case, c = a + ε(b - a) = εb + (1 - ε)aIn the second case, c = a + ε²(b - a) = ε²b + (1 - ε²)aSo, these give two different equations:1. ε²a + εb - c = 02. εa + ε²b - c = 0But the problem states a + ε²b + ε⁴c = 0 or a + ε⁴b + ε²c = 0.Since ε⁴ = ε, as ε³ = 1, so ε⁴ = ε.So, the given conditions are:1. a + ε²b + εc = 02. a + εb + ε²c = 0Comparing with what I have:From the first rotation, I have ε²a + εb - c = 0, which can be rewritten as ε²a + εb = c.If I multiply both sides by ε²:ε⁴a + ε³b = ε²cBut ε⁴ = ε, ε³ = 1, so:εa + b = ε²cRearranged: ε²c = εa + bMultiply both sides by ε²:ε⁴c = ε³a + ε²bBut ε⁴ = ε, ε³ = 1, so:εc = a + ε²bWhich gives a + ε²b - εc = 0Wait, that's similar to the given condition but with a negative sign on εc.Hmm, maybe I made a mistake in the algebra.Alternatively, let's consider the given condition a + ε²b + εc = 0.If I factor out ε² from the last two terms:a + ε²(b + ε²c) = 0But ε² * ε² = ε⁴ = ε, so:a + ε²b + εc = 0Wait, that doesn't seem helpful.Alternatively, maybe I need to express a in terms of b and c.From a + ε²b + εc = 0, we get a = -ε²b - εc.Similarly, from a + εb + ε²c = 0, we get a = -εb - ε²c.So, if a is expressed as a linear combination of b and c with coefficients involving ε and ε², that might represent the rotation.Alternatively, maybe I should consider the vectors from the origin. If a, b, c are vertices of an equilateral triangle, then the vectors a, b, c satisfy certain rotational properties.But I'm getting a bit stuck here. Maybe I should look at the converse. Suppose a + ε²b + εc = 0, then show that the triangle is equilateral.If a + ε²b + εc = 0, then a = -ε²b - εc.So, the vector a is a linear combination of b and c with coefficients ε² and ε.Since ε and ε² are complex numbers representing rotations, this might imply that the points are related by rotation, hence forming an equilateral triangle.But I need to make this more precise.Alternatively, maybe I can compute the distances between the points.If a + ε²b + εc = 0, then let's compute |a - b|², |b - c|², and |c - a|².But that might be complicated. Alternatively, consider that if a + ε²b + εc = 0, then multiplying both sides by ε:εa + ε³b + ε²c = 0But ε³ = 1, so:εa + b + ε²c = 0Which is the same as the other condition: a + εb + ε²c = 0.Wait, no, because if I multiply the original equation by ε, I get εa + b + ε²c = 0, which is the other condition.So, these two conditions are related by multiplication by ε.Therefore, if one holds, the other holds as well, but scaled by ε.But since ε is a root of unity, scaling by ε doesn't change the geometric configuration, just rotates it.So, essentially, both conditions are equivalent in terms of the geometric configuration.Therefore, if a + ε²b + εc = 0, then the triangle is equilateral, and vice versa.I think I'm getting closer. Maybe I should summarize:Given that ε is a primitive cube root of unity, the condition a + ε²b + εc = 0 (or a + εb + ε²c = 0) arises from the rotational symmetry of an equilateral triangle. Specifically, one vertex can be obtained by rotating another vertex around the third by 120 degrees, which corresponds to multiplication by ε or ε² in the complex plane. Therefore, the given conditions are equivalent to the triangle being equilateral.Now, moving on to part (b). It says: Prove that the points a, b, and c are the vertices of an equilateral triangle if and only if a² + b² + c² = ab + bc + ac.Hmm, this seems like a symmetric condition. I remember that for three numbers, a² + b² + c² = ab + bc + ac implies that the numbers are equal or form an equilateral triangle in some sense.Wait, actually, in real numbers, a² + b² + c² = ab + bc + ac implies that (a - b)² + (b - c)² + (c - a)² = 0, which means a = b = c. But in complex numbers, it's different because the squares can cancel out differently.But in the context of complex numbers representing points in the plane, this condition might relate to the distances between the points.Let me recall that for three points a, b, c in the complex plane, the condition a² + b² + c² = ab + bc + ac can be rewritten as:a² + b² + c² - ab - bc - ac = 0Which is equivalent to:(1/2)[(a - b)² + (b - c)² + (c - a)²] = 0But in complex numbers, the square of a complex number is not necessarily non-negative, so this doesn't directly imply that the differences are zero.However, if a, b, c form an equilateral triangle, then the distances between each pair are equal. So, |a - b| = |b - c| = |c - a|.But how does that relate to the given condition?Alternatively, maybe using the condition from part (a). If a + ε²b + εc = 0, then perhaps squaring both sides or taking magnitudes could lead to the condition a² + b² + c² = ab + bc + ac.Let me try that.From part (a), we have a + ε²b + εc = 0.Let me compute the magnitude squared of both sides:|a + ε²b + εc|² = 0Which implies:(a + ε²b + εc)(overline{a} + overline{ε²} overline{b} + overline{ε} overline{c}) = 0But since ε is a root of unity, overline{ε} = ε² and overline{ε²} = ε.So, expanding the product:aoverline{a} + aoverline{ε²}overline{b} + aoverline{ε}overline{c} + ε²boverline{a} + ε²boverline{ε²}overline{b} + ε²boverline{ε}overline{c} + εcoverline{a} + εcoverline{ε²}overline{b} + εcoverline{ε}overline{c} = 0Simplify each term:|a|² + a ε overline{b} + a ε² overline{c} + ε² overline{a} b + |b|² + ε² b ε overline{c} + ε overline{a} c + ε c overline{b} + |c|² = 0Wait, this is getting complicated. Maybe instead of taking magnitudes, I should consider multiplying the equation by its conjugate.Alternatively, maybe consider that if a + ε²b + εc = 0, then taking the conjugate:overline{a} + ε² overline{b} + ε overline{c} = 0But I'm not sure if that helps.Alternatively, maybe square both sides of the equation a + ε²b + εc = 0.But squaring in complex numbers isn't straightforward because it's not commutative. Wait, actually, in this case, since we're dealing with complex numbers, squaring would involve multiplying the expression by itself.So, (a + ε²b + εc)(a + ε²b + εc) = 0Expanding this:a² + a ε²b + a εc + ε²b a + (ε²b)² + ε²b εc + εc a + εc ε²b + (εc)² = 0Simplify each term:a² + a ε²b + a εc + ε²a b + ε⁴b² + ε³b c + εa c + ε³c b + ε²c² = 0Now, recall that ε³ = 1, so ε⁴ = ε, and ε³ = 1.So, substituting:a² + a ε²b + a εc + ε²a b + εb² + b c + εa c + c b + ε²c² = 0Combine like terms:a² + ε²c² + (a ε²b + ε²a b) + (a εc + εa c) + εb² + b c + c b = 0Factor out common terms:a² + ε²c² + ε²ab(1 + 1) + εac(1 + 1) + εb² + 2bc = 0Wait, no, let's see:a ε²b + ε²a b = 2 ε²abSimilarly, a εc + εa c = 2 εacSo, the equation becomes:a² + ε²c² + 2 ε²ab + 2 εac + εb² + 2bc = 0Hmm, not sure if this is helpful. Maybe I need a different approach.Alternatively, let's consider the given condition a² + b² + c² = ab + bc + ac.Let me rearrange it:a² + b² + c² - ab - bc - ac = 0Multiply both sides by 2:2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0Which can be written as:(a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ac + a²) = 0But that's:(a - b)² + (b - c)² + (c - a)² = 0Wait, but in complex numbers, the square of a complex number isn't necessarily non-negative, so this doesn't directly imply that each term is zero. However, if the sum is zero, it might imply some relationship between the terms.But in the context of points in the plane, if (a - b)² + (b - c)² + (c - a)² = 0, then each term must be zero because the squares of the distances are non-negative. Wait, but in complex numbers, the square of a complex number isn't a distance, it's another complex number.Wait, maybe I'm confusing real and complex numbers. In real numbers, (a - b)² is non-negative, but in complex numbers, (a - b)² can be any complex number.So, this approach might not work. Maybe I need to think differently.Alternatively, let's use the condition from part (a). If a + ε²b + εc = 0, then let's compute a² + b² + c².From a + ε²b + εc = 0, we can express a = -ε²b - εc.Then, a² = ( -ε²b - εc )² = ε⁴b² + 2 ε³b c + ε²c²But ε³ = 1, so ε⁴ = ε, and ε³ = 1.Thus, a² = εb² + 2b c + ε²c²Similarly, compute b² + c²:b² + c² = b² + c²So, a² + b² + c² = εb² + 2b c + ε²c² + b² + c² = (1 + ε)b² + (1 + ε²)c² + 2b cNow, compute ab + bc + ac.From a = -ε²b - εc,ab = (-ε²b - εc)b = -ε²b² - εb cSimilarly, ac = (-ε²b - εc)c = -ε²b c - εc²So, ab + bc + ac = (-ε²b² - εb c) + b c + (-ε²b c - εc²) = -ε²b² - εb c + b c - ε²b c - εc²Combine like terms:-ε²b² - εc² + (-εb c - ε²b c) + b cFactor out terms:-ε²b² - εc² - εb c(1 + ε) + b cBut 1 + ε + ε² = 0, so 1 + ε = -ε²Thus, -εb c(1 + ε) = -εb c(-ε²) = ε³b c = 1 * b c = b cSo, ab + bc + ac = -ε²b² - εc² + b c + b c = -ε²b² - εc² + 2b cNow, compare a² + b² + c² and ab + bc + ac:a² + b² + c² = (1 + ε)b² + (1 + ε²)c² + 2b cab + bc + ac = -ε²b² - εc² + 2b cSo, set a² + b² + c² = ab + bc + ac:(1 + ε)b² + (1 + ε²)c² + 2b c = -ε²b² - εc² + 2b cSubtract 2b c from both sides:(1 + ε)b² + (1 + ε²)c² = -ε²b² - εc²Bring all terms to one side:(1 + ε + ε²)b² + (1 + ε² + ε)c² = 0But 1 + ε + ε² = 0, so:0 * b² + 0 * c² = 0Which is 0 = 0.So, the condition a² + b² + c² = ab + bc + ac holds if a + ε²b + εc = 0.Therefore, the two conditions are equivalent.Conversely, if a² + b² + c² = ab + bc + ac, then from the above steps, we can reverse the process and show that a + ε²b + εc = 0 or a + εb + ε²c = 0, which implies the triangle is equilateral.So, putting it all together, part (b) is proven by using the condition from part (a) and manipulating it to show that a² + b² + c² = ab + bc + ac.

💡First, I need to determine the total number of balls in the bag. There are 5 green balls, 8 white balls, 4 blue balls, and 3 red balls, which adds up to 20 balls in total.Next, I'll calculate the total number of ways to draw 4 balls from these 20 balls using the combination formula ( C(n, k) = frac{n!}{k!(n - k)!} ). This gives me ( C(20, 4) = 4845 ).Now, I'll consider each color separately to find the number of ways to draw 4 balls of the same color:1. **Green Balls**: There are 5 green balls, so the number of ways to choose 4 green balls is ( C(5, 4) = 5 ).2. **White Balls**: There are 8 white balls, so the number of ways to choose 4 white balls is ( C(8, 4) = 70 ).3. **Blue Balls**: There are 4 blue balls, so the number of ways to choose 4 blue balls is ( C(4, 4) = 1 ).4. **Red Balls**: There are only 3 red balls, so it's impossible to draw 4 red balls, resulting in 0 ways.Adding up the number of ways for each color gives me ( 5 + 70 + 1 = 76 ).Finally, I'll calculate the probability by dividing the total number of favorable outcomes by the total number of possible outcomes: ( frac{76}{4845} ). This fraction cannot be simplified further, so the probability remains ( frac{76}{4845} ).

💡Alright, so I'm trying to figure out how many coins are in each pile that Roger has. He has 3 piles of quarters and 3 piles of dimes, and altogether, there are 42 coins. Hmm, okay, let's break this down.First, I need to understand what exactly is being asked. Roger has two types of coins: quarters and dimes. For each type, he has 3 piles. So, there are 3 piles of quarters and 3 piles of dimes. The total number of coins from both quarters and dimes is 42.Now, the question is asking how many coins are in each pile. That means I need to find out how many quarters are in each of the 3 quarter piles and how many dimes are in each of the 3 dime piles. But wait, the problem says "each pile," which could mean that all piles have the same number of coins, regardless of whether they're quarters or dimes. Or it could mean that each type of pile (quarters and dimes) has the same number of coins within their own type.Let me think about this. If all piles have the same number of coins, then each pile, whether it's a quarter pile or a dime pile, has the same number of coins. So, there are 3 quarter piles and 3 dime piles, making a total of 6 piles. If each pile has the same number of coins, then the total number of coins would be 6 times the number of coins in each pile.But the total number of coins is 42, so if I let x be the number of coins in each pile, then 6x = 42. Solving for x, I divide both sides by 6, and I get x = 7. So, each pile has 7 coins. That seems straightforward.But wait, maybe I'm oversimplifying. The problem says "each pile had a certain number of coins in it." This could imply that each pile might have a different number of coins, not necessarily the same. But then, the problem also says "how many coins were in each pile," which could mean that all piles have the same number of coins.I think the key here is to see if there's a unique solution or if there are multiple possibilities. If all piles have the same number of coins, then it's 7 coins per pile. But if the number of coins in each pile can vary, then there might be multiple solutions.Let me try to set up an equation to model this. Let's say that each quarter pile has q coins, and each dime pile has d coins. Since there are 3 quarter piles and 3 dime piles, the total number of coins is 3q + 3d = 42. Simplifying this equation, I can factor out the 3: 3(q + d) = 42. Dividing both sides by 3, I get q + d = 14.So, the sum of the number of coins in each quarter pile and each dime pile is 14. This means that there are multiple possible combinations of q and d that satisfy this equation. For example, if q = 5, then d = 9; if q = 6, then d = 8; and so on.But the problem asks "how many coins were in each pile." If it's asking for the number of coins in each pile regardless of type, then the answer would be 7 coins per pile, assuming all piles have the same number of coins. However, if it's asking for the number of coins in each type of pile (quarters and dimes), then there are multiple possible answers depending on the values of q and d.Given that the problem mentions both quarters and dimes, it's likely that it's asking for the number of coins in each type of pile. Therefore, the answer would be that each quarter pile has q coins and each dime pile has d coins, where q + d = 14. Without additional information, we can't determine the exact number of coins in each type of pile, only that their sum is 14.Wait, but the initial assumption that all piles have the same number of coins led to a unique solution of 7 coins per pile. Maybe that's the intended answer, assuming uniformity across all piles. However, the problem doesn't explicitly state that all piles have the same number of coins, only that each pile has a certain number of coins.So, to reconcile this, perhaps the problem expects us to assume that each pile has the same number of coins, leading to 7 coins per pile. Alternatively, if we consider that each type of pile (quarters and dimes) has the same number of coins within their own type, then we can have different numbers for quarters and dimes.Given the ambiguity, I think the most straightforward answer is that each pile has 7 coins, assuming uniformity across all piles. Therefore, each of the 3 quarter piles has 7 quarters, and each of the 3 dime piles has 7 dimes, totaling 42 coins.But to be thorough, let's explore the other possibility where each type of pile has the same number of coins within their own type. Let's say each quarter pile has q coins and each dime pile has d coins, with q possibly different from d. Then, 3q + 3d = 42, which simplifies to q + d = 14. So, q and d can be any pair of numbers that add up to 14. For example:- q = 5, d = 9- q = 6, d = 8- q = 7, d = 7- q = 8, d = 6- q = 9, d = 5And so on. Therefore, without additional information, we can't determine the exact number of coins in each type of pile, only that their sum is 14.However, since the problem asks "how many coins were in each pile," and not specifying quarters or dimes, it's likely that it's asking for the number of coins in each pile regardless of type. Therefore, assuming uniformity, each pile has 7 coins.In conclusion, the most reasonable answer is that each pile has 7 coins.

💡Okay, so I have this problem here where two real numbers, x and y, satisfy two equations: x minus y equals 3, and x cubed minus y cubed equals 27. I need to find the product of x and y, which is xy. Hmm, let me think about how to approach this.First, I remember that when dealing with equations involving x and y, substitution is often a good method. Since the first equation is x - y = 3, maybe I can solve for one variable in terms of the other and substitute it into the second equation. That should help me find the values of x and y, and then I can compute xy.Alright, so starting with the first equation: x - y = 3. Let me solve for x in terms of y. If I add y to both sides, I get x = y + 3. That seems straightforward. Now, I can substitute this expression for x into the second equation, which is x³ - y³ = 27. Substituting x = y + 3 into the second equation gives me (y + 3)³ - y³ = 27. Now, I need to expand (y + 3)³. I recall that the formula for expanding (a + b)³ is a³ + 3a²b + 3ab² + b³. Applying that here, (y + 3)³ would be y³ + 3y²*3 + 3y*3² + 3³. Let me compute each term step by step.First term: y³. Second term: 3y²*3 is 9y². Third term: 3y*9 is 27y. Fourth term: 27. So putting it all together, (y + 3)³ is y³ + 9y² + 27y + 27. Now, subtracting y³ from this, the equation becomes (y³ + 9y² + 27y + 27) - y³ = 27. Simplifying this, the y³ terms cancel out, leaving 9y² + 27y + 27 = 27. Hmm, okay, so now I have 9y² + 27y + 27 = 27. Let me subtract 27 from both sides to simplify further. That gives me 9y² + 27y = 0. Looking at this equation, I can factor out a common term. Both terms have a factor of 9y, so factoring that out gives me 9y(y + 3) = 0. Wait, hold on, 9y² + 27y is equal to 9y(y + 3). So, 9y(y + 3) = 0. Setting each factor equal to zero gives me the possible solutions for y. So, 9y = 0 implies y = 0, and y + 3 = 0 implies y = -3. Hmm, wait, but in the original equation, x - y = 3, so if y is -3, then x would be 0. Let me check if these solutions satisfy the second equation, x³ - y³ = 27.First, let's take y = 0. Then x = y + 3 = 0 + 3 = 3. So, x = 3 and y = 0. Plugging into x³ - y³, we get 3³ - 0³ = 27 - 0 = 27, which matches the second equation. So that's a valid solution.Next, let's take y = -3. Then x = y + 3 = -3 + 3 = 0. So, x = 0 and y = -3. Plugging into x³ - y³, we get 0³ - (-3)³ = 0 - (-27) = 27, which also matches the second equation. So, both solutions are valid.Wait a second, so both y = 0 and y = -3 are solutions? That means there are two possible pairs for (x, y): (3, 0) and (0, -3). Therefore, the product xy can be either 3*0 = 0 or 0*(-3) = 0. So, in both cases, xy equals 0.But hold on, I think I made a mistake earlier when factoring. Let me double-check that step. The equation after substitution was 9y² + 27y = 0. Factoring out 9y gives 9y(y + 3) = 0, which leads to y = 0 or y = -3. That seems correct.Wait, but in the initial problem statement, it says "two real numbers x and y." So, both solutions are valid, and in both cases, the product xy is 0. Therefore, the answer should be 0.But just to be thorough, let me verify both solutions again.First solution: x = 3, y = 0. Check x - y = 3 - 0 = 3, which satisfies the first equation. x³ - y³ = 27 - 0 = 27, which satisfies the second equation.Second solution: x = 0, y = -3. Check x - y = 0 - (-3) = 3, which satisfies the first equation. x³ - y³ = 0 - (-27) = 27, which satisfies the second equation.So, both solutions are correct, and in both cases, the product xy is 0. Therefore, the value of xy is 0.Wait, but I just realized something. When I factored 9y² + 27y = 0, I got 9y(y + 3) = 0, leading to y = 0 or y = -3. But in the original problem, the second equation is x³ - y³ = 27. If I plug in x = 0 and y = -3, I get 0 - (-27) = 27, which is correct. Similarly, x = 3 and y = 0 gives 27 - 0 = 27, which is also correct.Therefore, both solutions are valid, and in both cases, the product xy is 0. So, the answer is 0.But just to make sure I didn't miss any other possible solutions, let me think about whether there could be more solutions. The equation 9y² + 27y = 0 is a quadratic equation, which can have at most two real solutions. Since we found both y = 0 and y = -3, there are no other real solutions. Therefore, these are the only possible pairs (x, y) that satisfy both equations.Thus, the product xy is 0 in both cases, so the answer is 0.

💡Okay, so I have this matrix:[begin{pmatrix} a & 4 -12 & d end{pmatrix}]And I need to figure out if it can be its own inverse. Hmm, I remember that if a matrix is its own inverse, then multiplying it by itself should give the identity matrix. So, let me write that down:[begin{pmatrix} a & 4 -12 & d end{pmatrix} times begin{pmatrix} a & 4 -12 & d end{pmatrix} = begin{pmatrix} 1 & 0 0 & 1 end{pmatrix}]Alright, now I need to perform this matrix multiplication. Let me recall how matrix multiplication works. Each element of the resulting matrix is the dot product of the corresponding row of the first matrix and column of the second matrix.So, let's compute each element one by one.First element (top-left):[a times a + 4 times (-12) = a^2 - 48]Second element (top-right):[a times 4 + 4 times d = 4a + 4d]Third element (bottom-left):[-12 times a + d times (-12) = -12a - 12d]Fourth element (bottom-right):[-12 times 4 + d times d = -48 + d^2]So, putting it all together, the product matrix is:[begin{pmatrix} a^2 - 48 & 4a + 4d -12a - 12d & d^2 - 48 end{pmatrix}]And this should equal the identity matrix:[begin{pmatrix} 1 & 0 0 & 1 end{pmatrix}]Therefore, we can set up equations for each corresponding element:1. ( a^2 - 48 = 1 )2. ( 4a + 4d = 0 )3. ( -12a - 12d = 0 )4. ( d^2 - 48 = 1 )Let me look at these equations one by one.Starting with equation 1:( a^2 - 48 = 1 )Adding 48 to both sides:( a^2 = 49 )So, ( a = sqrt{49} ) or ( a = -sqrt{49} ), which means ( a = 7 ) or ( a = -7 ).Similarly, equation 4 is:( d^2 - 48 = 1 )Adding 48 to both sides:( d^2 = 49 )So, ( d = 7 ) or ( d = -7 ).Now, looking at equations 2 and 3:Equation 2: ( 4a + 4d = 0 )Equation 3: ( -12a - 12d = 0 )I notice that equation 3 is just equation 2 multiplied by -3. So, they are not independent equations. Therefore, we can just consider equation 2.Simplifying equation 2:( 4a + 4d = 0 )Divide both sides by 4:( a + d = 0 )So, ( d = -a )Now, since ( a ) can be 7 or -7, let's find the corresponding ( d ).If ( a = 7 ), then ( d = -7 ).If ( a = -7 ), then ( d = 7 ).So, the possible pairs are (7, -7) and (-7, 7).Let me double-check to make sure these satisfy all the equations.First, for (7, -7):Equation 1: ( 7^2 - 48 = 49 - 48 = 1 ) ✔️Equation 2: ( 4*7 + 4*(-7) = 28 - 28 = 0 ) ✔️Equation 3: ( -12*7 -12*(-7) = -84 + 84 = 0 ) ✔️Equation 4: ( (-7)^2 - 48 = 49 - 48 = 1 ) ✔️Now, for (-7, 7):Equation 1: ( (-7)^2 - 48 = 49 - 48 = 1 ) ✔️Equation 2: ( 4*(-7) + 4*7 = -28 + 28 = 0 ) ✔️Equation 3: ( -12*(-7) -12*7 = 84 - 84 = 0 ) ✔️Equation 4: ( 7^2 - 48 = 49 - 48 = 1 ) ✔️Both pairs satisfy all the equations. So, there are two possible pairs: (7, -7) and (-7, 7).I think that's all. So, the number of possible pairs is 2.**Final Answer**The number of possible pairs ((a, d)) is boxed{2}.

💡First, I need to find the product of the two functions ( f(x) = frac{x - 3}{x + 3} ) and ( g(x) = x + 3 ).I'll start by writing out the product:[f(x) cdot g(x) = frac{x - 3}{x + 3} cdot (x + 3)]Next, I notice that the ( x + 3 ) in the denominator and the ( x + 3 ) being multiplied in the numerator can cancel each other out, provided that ( x neq -3 ) to avoid division by zero.After canceling, the expression simplifies to:[f(x) cdot g(x) = x - 3]It's important to specify the domain of the original functions to ensure the simplification is valid. The original function ( f(x) ) is undefined at ( x = -3 ), so the domain of the product is all real numbers except ( x = -3 ).Therefore, the final simplified product is ( x - 3 ) with the domain ( x in (-infty, -3) cup (-3, +infty) ).