How do you factor x^3+27=0?

Feb 18, 2017

$\left(x + 3\right) \left({x}^{2} - 3 x + 9\right) = 0$

Explanation:

We have a standard polynomial on LHS of the form ${a}^{3} + {b}^{3}$, whose one factor is $\left(a + b\right)$. So one factor is $\left(x + 3\right)$

${x}^{3} + 27 = 0$

$\Leftrightarrow {x}^{3} + 3 {x}^{2} - 3 {x}^{2} - 9 x + 9 x + 27 = 0$

or ${x}^{2} \left(x + 3\right) - 3 x \left(x + 3\right) + 9 \left(x + 3\right) = 0$

or $\left(x + 3\right) \left({x}^{2} - 3 x + 9\right) = 0$