Question

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

Answer 1

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

$=(3x{)}^3+y^3+z^3-9xyz$

$=(3x{)}^3+y^3+z^3-3\times 3x\times y\times z$

using identity

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

Putting $a=3x,b=y,c=z$

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