Question

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

Answer 1

Factorization

The given equation is,

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

We know that,

${\mathit{a}}^{\mathbf{3}}\mathbf{+}{\mathit{b}}^{\mathbf{3}}\mathbf{+}{\mathit{c}}^{\mathbf{3}}\mathbf{-}\mathbf{3}\mathit{a}\mathit{b}\mathit{c}\mathbf{=}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathit{c}\mathbf{)}\mathbf{(}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{+}{\mathit{c}}^{\mathbf{2}}\mathbf{-}\mathit{a}\mathit{b}\mathbf{-}\mathit{b}\mathit{c}\mathbf{-}\mathit{c}\mathit{a}\mathbf{)}$

$\therefore 27x^3+y^3+z^3-9xyz=(3x+y+z)(9x^2+y^2+z^2-3xy-yz-3xz)$

Hence, the result of the factorization of the given equation is $(3x+y+z)(9x^2+y^2+z^2-3xy-yz-3xz)$