# How do you factor b^3 + 27?

Feb 3, 2016

${b}^{3} + 27 = \left(b + 3\right) \left({b}^{2} - 3 b + 9\right)$

#### Explanation:

${b}^{3} + 27$ is an example of a sum of cubes, where $a = b$ and $b = 3.$. Rewrite the expression as ${\left(b\right)}^{3} + {\left(3\right)}^{3}$.

To solve the sum of cubes, use the equation ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$.

Substitute the values for $a \mathmr{and} b$ into the equation.

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

Simplify.

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