Question:
If \[ x+y+z=8 \] and \[ xy+yz+zx=20 \] find:
\[ x^3+y^3+z^3-3xyz \]
Solution:
Using identity:
\[ x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \]
First find \[ x^2+y^2+z^2 \]
Using identity:
\[ (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \]
Substituting the given values:
\[ 8^2 = x^2+y^2+z^2+2(20) \]
\[ 64 = x^2+y^2+z^2+40 \]
\[ x^2+y^2+z^2 = 24 \]
Now,
\[ x^3+y^3+z^3-3xyz \]
\[ = 8(24-20) \]
\[ = 8\times4 \]
\[ =32 \]