# How do you factor 8x^3y^6 + 27?

Apr 22, 2015

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
(this is either something you already know or you can derive it by synthetic division/multiplication)

$8 {x}^{3} {y}^{6} + 27 = {\left(2 x {y}^{2}\right)}^{3} + {3}^{3}$

So this can be factored as
$\left(2 x {y}^{2} + 3\right) \left({\left(2 x {y}^{2}\right)}^{2} - \left(2 x {y}^{2}\right) \left(3\right) + {3}^{2}\right)$
or
$\left(2 x {y}^{2} + 3\right) \left(4 {x}^{2} {y}^{4} - 6 x {y}^{2} + 9\right)$

You might be tempted to try to factor the second part of this but if you consider the formula for roots of a quadratic:
$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

we have the square root of a negative value; so there are no Real factors available.