# How do you factor 8k^3 + 1?

##### 1 Answer
Feb 11, 2016

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

#### Explanation:

This is the factorization of perfect cubes.

The pattern for factoring perfect cubes is
$\left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
or
$\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
The pattern is dependent upon the sign between the cubes.

a = the cubic factor of the first term
b = the cubic root of the second term

(first term, second term) (first term squared, first term x second term, second term squared.

The pattern of the pattern of the signs follows the SOAP rule
S - Same Sign
O - Opposite Sign
AP - Always Positive

(Same Sign) (Opposite Sign, Always Positive)

$8 {k}^{3} + 1$
The cubic factor of $8 {k}^{3}$ is 2k
The cubic factor of 1 is 1

$a = 2 k$
$b = 1$

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