Question

Prove that: ${(x+y)}^3+{(y+z)}^3+{(z+x)}^3-3(x+y)(y+z)(z+x)=2(x^3+y^3+z^3-3xyz)$

Answer 1

$\to$ As $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$=\frac{1}{2}(a+b+c)\left(\right(a-b{)}^2+(b-c{)}^2+(a-c{)}^2)$

So LHS becomes,

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

$=(x+y+z)\times 2(x^2+y^2+z^2-xy-yz-xz)$

$=2(x^3+y^3+z^3-3xyz)=RHS$