# How do you factor 27x^3+64?

Apr 12, 2015

When factorising the sum of cubes, we use the formula
${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$.

In this case of $27 x 3 + 64$,
$27 {x}^{3} = {a}^{3}$
$64 = {b}^{3}$

Find $a$:
$27 {x}^{3} = {a}^{3}$
$\sqrt[3]{27 {x}^{3}} = \sqrt[3]{{a}^{3}}$
$3 x = a$

Find $b$:
$64 = {b}^{3}$
$\sqrt[3]{64} = \sqrt[3]{{b}^{3}}$
$4 = b$

Substitute $a = 3 x$ and $b = 4$ into $\left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

$\left(3 x + 4\right) \left({\left(3 x\right)}^{2} - \left(3 x \times 4\right) + {4}^{2}\right)$

= $\left(3 x + 4\right) \left(9 {x}^{2} - 12 x + 16\right)$

$\left(3 x + 4\right) \left(9 {x}^{2} - 12 x + 16\right)$ is the factorised form of $27 x 3 + 64$