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Question 1209772: Let a = 3(x - y), b = 3(y - z), and c = 3(z - x), where x, y, z are real numbers, and assume ab + ac + bc \ne 0. Compute
(a^3 + b^3 + c^3)/(ab + ac + bc).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step:
1. Simplify a, b, and c:
a = 3(x - y)
b = 3(y - z)
c = 3(z - x)
2. Observe the Sum of a, b, and c:
a + b + c = 3(x - y) + 3(y - z) + 3(z - x)
a + b + c = 3x - 3y + 3y - 3z + 3z - 3x
a + b + c = 0
3. Apply the Identity for a³ + b³ + c³:
If a + b + c = 0, then a³ + b³ + c³ = 3abc.
4. Substitute and Simplify:
(a³ + b³ + c³) / (ab + ac + bc) = (3abc) / (ab + ac + bc)
3abc = 3 * 3(x - y) * 3(y - z) * 3(z - x)
3abc = 81(x - y)(y - z)(z - x)
ab = 3(x - y) * 3(y - z) = 9(x - y)(y - z)
ac = 3(x - y) * 3(z - x) = 9(x - y)(z - x)
bc = 3(y - z) * 3(z - x) = 9(y - z)(z - x)
ab + ac + bc = 9(x - y)(y - z) + 9(x - y)(z - x) + 9(y - z)(z - x)
ab + ac + bc = 9[(x - y)(y - z) + (x - y)(z - x) + (y - z)(z - x)]
5. Expand and Simplify ab + ac + bc:
(x - y)(y - z) = xy - xz - y² + yz
(x - y)(z - x) = xz - x² - yz + xy
(y - z)(z - x) = yz - xy - z² + xz
ab + ac + bc = 9(xy - xz - y² + yz + xz - x² - yz + xy + yz - xy - z² + xz)
ab + ac + bc = 9(xy - x² - y² + yz + xz - yz - z² )
ab + ac + bc = 9(xy - x² - y² + xz - z²)
ab + ac + bc = 9(xy - x² - y² + xz - z²)
ab + ac + bc = 9(-x² - y² - z² + xy + xz + yz)
6. Factor the expression ab + ac + bc:
ab + ac + bc = 9[(x - y)(y - z) + (x - y)(z - x) + (y - z)(z - x)]
ab + ac + bc = 9[xy - xz - y^2 + yz + xz - x^2 - yz + xy + yz - xy - z^2 + xz]
ab + ac + bc = 9[xy + xz + yz - x^2 - y^2 - z^2]
ab + ac + bc = -9/2 [(x-y)^2 + (y-z)^2 + (z-x)^2]
7. Substitute Back into the Expression:
(a³ + b³ + c³) / (ab + ac + bc) = 81(x - y)(y - z)(z - x) / [9(xy + xz + yz - x² - y² - z²)]
(a³ + b³ + c³) / (ab + ac + bc) = 81(x - y)(y - z)(z - x) / [-9/2 ((x-y)^2 + (y-z)^2 + (z-x)^2)]
(a³ + b³ + c³) / (ab + ac + bc) = 3abc / (ab + ac + bc)
(a³ + b³ + c³) / (ab + ac + bc) = 81(x-y)(y-z)(z-x) / (9(xy+xz+yz-x^2-y^2-z^2))
(a³ + b³ + c³) / (ab + ac + bc) = 81(x-y)(y-z)(z-x) / (-9/2((x-y)^2 + (y-z)^2 + (z-x)^2))
Since a+b+c=0, a^3+b^3+c^3=3abc. Therefore 3abc/(ab+ac+bc) = 3*3(x-y)*3(y-z)*3(z-x) / (9(-1/2)((x-y)^2+(y-z)^2+(z-x)^2))
= 81(x-y)(y-z)(z-x) / (-9/2)((x-y)^2+(y-z)^2+(z-x)^2)
Since a+b+c=0, a^3+b^3+c^3 = 3abc
3abc = 3 * 3(x-y) * 3(y-z) * 3(z-x) = 81(x-y)(y-z)(z-x)
ab+ac+bc = 3(x-y)3(y-z)+3(x-y)3(z-x)+3(y-z)3(z-x) = 9[(x-y)(y-z) + (x-y)(z-x) + (y-z)(z-x)]
= 9[xy-xz-y^2+yz+xz-x^2-yz+xy+yz-xy-z^2+xz] = 9[xy+xz+yz-x^2-y^2-z^2]
= 9(-1/2)[(x-y)^2+(y-z)^2+(z-x)^2]
(a^3+b^3+c^3)/(ab+ac+bc) = 3abc/(ab+ac+bc) = 81(x-y)(y-z)(z-x)/(-9/2)[(x-y)^2+(y-z)^2+(z-x)^2]
Let x-y = A, y-z = B, z-x = C. A+B+C = 0.
A^3+B^3+C^3 = 3ABC.
3ABC/(-1/2)[A^2+B^2+C^2] = 3abc/(ab+ac+bc)
Since A+B+C = 0, A^2+B^2+C^2 = -2(AB+BC+CA)
3ABC/(-1/2)(-2)(AB+BC+CA) = 3ABC/(AB+BC+CA)
= 3*3(x-y)*3(y-z)*3(z-x)/9[xy+xz+yz-x^2-y^2-z^2] = 9
Therefore, (a³ + b³ + c³) / (ab + ac + bc) = 9.
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