If x, y, z satisfy:
x + y + z = 12,
1/x + 1/y + 1/z = 2, and
x3 + y3 + z3 = -480,
find x2y + xy2 + x2z + xz2 + y2z + yz2.
First I will take the reciprocal of the second equation and do some simplifying to get xyz/(xy+yz+xz) = 1/2. Which can also be expressed as 2*xyz = xy+yz+xz
S1 = P1 = 12
2*P3 = P2
S3 = -480
S2-S1*P1+2*P2 = 0
S3-S2*P1+S1*P2-3*P3 = 0
Substitute to get
S2 + 4*P3 = 144
-12*S2 + *P3 = 480
Then S2 = 16 and P3 = 32
Now the goal expression; start with the expansion (x+y+z)^3 = [x^3+y^3+z^3] + 3*[x^2*y + x*y^2 + x^2*z + x*z^2 + y^2*z + y*z^2] + 6*xyz
Then x^2*y + x*y^2 + x^2*z + x*z^2 + y^2*z + y*z^2 = (S1^3 - S3 - 6*P3)/3
= (12^3 + 480 - 6*32)/3 = 672.
Solution
