# How do you factor b^3-64c^3?

Jan 1, 2016

Apply the difference of cubes formula to see that
${b}^{3} - 64 {c}^{3} = \left(b - 4 c\right) \left({b}^{2} + 4 b c + 16 {c}^{2}\right)$

#### Explanation:

The difference of cubes formula states that

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

(Verify this by multiplying out the right hand side)

Then, for the given expression, we have

${b}^{3} - 64 {c}^{3} = {b}^{3} - {\left(4 c\right)}^{3} = \left(b - 4 c\right) \left({b}^{2} + 4 b c + 16 {c}^{2}\right)$

Jan 1, 2016

${b}^{3} - 64 {c}^{3} = \left(b - 4 c\right) \left({b}^{2} + 4 b c + 16 {c}^{2}\right)$

#### Explanation:

${b}^{3} - 64 {c}^{3}$ is a difference of cubes . The formula for factoring a difference of cubes is $\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$, where $a = b$ and $b = 4 c$ in this case.

Substitute the values for $a \mathmr{and} b$ into the equation.

$\left(b - 4 c\right) \left({b}^{2} + b \times 4 c + {\left(4 c\right)}^{2}\right)$

Simplify.

$\left(b - 4 c\right) \left({b}^{2} + 4 b c + 16 {c}^{2}\right)$