# How do you factor (2x+1)^3 + (2y)^3?

##### 2 Answers
May 28, 2015

For any numbers a and b, ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

So, this expression becomes:

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

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

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

These 2 numbers look like they can be factored, but they can't because x and y are different variables.

May 28, 2015

You can use the identity for a sum of cubes:

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

substituting $a = 2 x + 1$ and $b = 2 y$ to get:

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

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

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

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