Question

Factorise: $27x^3+y^3+z^3-9xyz$

Answer 1

Given: $27x^3+y^3+z^3-9xyz$
$=(3x{)}^3+(y{)}^3+(z{)}^3-3(3x\left)\right(y\left)\right(z)$
Using the identity, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, where,

$a=3x,b=y$ and $c=z$, we get,

$27x^3+y^3+z^3-9xyz=(3x+y+z)\left(\right(3x{)}^2+y^2+z^2-3xy-yz-3xz)$

$=(3x+y+z)(9x^2+y^2+z^2-3xy-yz-3zx)$

Therefore,$27x^3+y^3+z^3-9xyz=(3x+y+z)(9x^2+y^2+z^2-3xy-yz-3zx)$