# How do you factor 27(K^3)-8?

May 30, 2016

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

#### Explanation:

This is an example of the difference of two cubes.
Look for the pattern here:

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

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

Hint: There are 2 brackets in the answer. The first is easy to find.
The second bracket is formed from the first bracket.

Using the same pattern...

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