# How do you factor 5u^3-40(x+y)^3?

May 15, 2016

$5 {u}^{3} - 40 {\left(x + y\right)}^{3} = 5 \left(u - 2 x - 2 y\right) \left({u}^{2} + 2 u x + 2 u y + 4 {x}^{2} + 8 x y + 4 {y}^{2}\right)$

#### Explanation:

We will use the difference of cubes identity:

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

with $a = u$ and $b = 2 \left(x + y\right)$

First separate out the common scalar factor $5$:

$5 {u}^{3} - 40 {\left(x + y\right)}^{3}$

$= 5 \left({u}^{3} - 8 {\left(x + y\right)}^{3}\right)$

$= 5 \left({u}^{3} - {\left(2 \left(x + y\right)\right)}^{3}\right)$

$= 5 \left(u - 2 \left(x + y\right)\right) \left({u}^{2} + u \left(2 \left(x + y\right)\right) + {\left(2 \left(x + y\right)\right)}^{2}\right)$

$= 5 \left(u - 2 x - 2 y\right) \left({u}^{2} + 2 u x + 2 u y + 4 {x}^{2} + 8 x y + 4 {y}^{2}\right)$