How do you factor 8u^3+27?

Jan 5, 2016

$8 {u}^{3} + 27 = \left(2 u + 3\right) \left(4 {u}^{2} - 6 u + 9\right)$

Explanation:

$8 {u}^{3} + 27$

Since $8 {u}^{3}$ and $27$ are cubes, we can rewrite the expression as ${\left(2 u\right)}^{3} + {\left(3\right)}^{3}$.

This a sum of two cubes with the form ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$, where $a = 2 u$ and $b = 3$.

Substitute the values for $a$ and $b$ into the equation.

${\left(2 u\right)}^{3} + {\left(3\right)}^{3} = \left(2 u + 3\right) {\left(2 u\right)}^{2} - \left(2 u\right) \left(3\right) + {\left(3\right)}^{2}$

Simplify.

${\left(2 u\right)}^{3} + {\left(3\right)}^{3} = \left(2 u + 3\right) \left(4 {u}^{2} - 6 u + 9\right)$