Question

If $3x+y+z=0$ show that $27x^3+y^3+z^3=9xyz$

Answer 1

We know that,

If $x+y+z=0$

then,

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

$\Rightarrow x^3+y^3+z^3-3xyz=0\times (x^2+y^2+z^2-xy-yz-zx)$

$\Rightarrow x^3+y^3+z^3-3xyz=0$

Now, given that

If $3x+y+z=0$

then,

${\left(3x\right)}^3+y^3+z^3-3\times 3xyz=(3x+y+z)({\left(3x\right)}^2+y^2+z^2-xy-yz-zx)$

$\Rightarrow 27x^3+y^3+z^3-9xyz=0\times (9x^2+y^2+z^2-xy-yz-zx)$

$\Rightarrow x^3+y^3+z^3=9xyz$

Hence, this is the answer.