Question

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

Answer 1

$27x^3+y^3+z^3-9xyz$ $=$ $(3x{)}^3+(y{)}^3+(z{)}^3-3(3a\left)\right(y\left)\right(z)$

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

Comparing both sides, we get

$\therefore (3x{)}^3+(y{)}^3+(z{)}^3-3(3x\left)\right(y\left)\right(z)$

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

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