Question

\(\dfrac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\)

Answer 1

\(\dfrac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\)

\(=\dfrac{\left(x+y\right)^3+z^3-\left(3x^2y+3xy^2+3xyz\right)}{x^2-2xy+y^2+z^2-2xz+z^2+y^2-2yz+z^2}\)

\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz-z^2\right)-3xy\left(x+y+z\right)}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)

\(=\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)}{2\left(x^2+y^2+z^2-xy-xz-yz\right)}\)

\(=\dfrac{x+y+z}{2}\)