# How do you factor 54b^3+16?

Aug 6, 2016

$54 {b}^{3} + 16 = 2 \left(3 b + 2\right) \left(9 {b}^{2} - 6 b + 4\right)$

#### Explanation:

The sum of cubes identity can be written:

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

Note that both $54$ and $16$ are divisible by $2$, so we find:

$54 {b}^{3} + 16$

$= 2 \left(27 {b}^{3} + 8\right)$

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

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

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