Question

Verify that $x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$.

Answer 1

we know that ,

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

taking RHS of above equation

$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

multiply 2 divided by 2

$\frac{1}{2}(x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-zx)$

now spliting the terms & rearranging

$\frac{1}{2}(x+y+z)(x^2+y^2-2xy+z^2+y^2-2yz+x^2+z^2-2xz)$

we know that

$[x^2+y^2-2xy=(x-y{)}^2]$ unity this in above equation we have

$\frac{1}{2}(x+y+z)\left[\right(x-y{)}^2+(y-t{)}^2+(z-x{)}^2]$

$\to x^3+y^3+z^3=\frac{1}{2}(x+y+z)\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$

hence proved.