# How do you factor 1000x^3 + 216 ?

Dec 31, 2015

Use the sum of cubes identity to get:

$1000 {x}^{3} + 216$

$= \left(10 x + 6\right) \left(100 {x}^{2} - 60 x + 36\right)$

$= 8 \left(5 x + 3\right) \left(25 {x}^{2} - 15 x + 9\right)$

#### Explanation:

Both $1000 {x}^{3} = {\left(10 x\right)}^{3}$ and $216 = {6}^{3}$ are perfect cubes, so we can use the sum of cubes identity:

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

with $a = 10 x$ and $b = 6$ as follows:

$1000 {x}^{3} + 216$

$= {\left(10 x\right)}^{3} + {6}^{3}$

=(10x+6)((10x)^2-(10x)(6)+6^2

$= \left(10 x + 6\right) \left(100 {x}^{2} - 60 x + 36\right)$

Alternatively, separate out the common scalar factor $8$, to get:

$= 8 \left(5 x + 3\right) \left(25 {x}^{2} - 15 x + 9\right)$