# How do you factor completely 8x^3+1?

Dec 30, 2015

Use the sum of cubes identity to find:

$8 {x}^{3} + 1 = \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right)$

#### Explanation:

The sum of cubes identity can be written:

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

Use this with $a = 2 x$ and $b = 1$ as follows:

$8 {x}^{3} + 1$

$= {\left(2 x\right)}^{3} + {1}^{3}$

$= \left(2 x + 1\right) \left({\left(2 x\right)}^{2} - \left(2 x\right) \left(1\right) + {1}^{2}\right)$

$= \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right)$