Question

How do you factor completely $\left(x^3-1\right)$?

Answer 1

Factorise the given expression using algebraic identity

According to the algebraic identities,

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

So, the given expression can be factorised as follows,

$x^3-1={\left(x\right)}^3-{\left(1\right)}^3$

$\Rightarrow x^3-1=\left(x-1\right)\left[{\left(x\right)}^2+x\times 1+{\left(1\right)}^2\right]$

$\Rightarrow x^3-1=\left(x-1\right)\left[x^2+x+1\right]$

Hence, the factors of $\left(x^3-1\right)$ obtained by using algebraic identity is $\left(x-1\right)\left(x^2+x+1\right)$.