# How do you factor completely 64 + a^3 ?

Jul 13, 2016

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

#### Explanation:

The expression ${a}^{3} + 64 = {a}^{3} + {4}^{3}$ is the sum of two cubes (3rd powers). In general, if you have an arbitrary sum of two cubes, say ${x}^{3} + {y}^{3}$, it can be factored as:

${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$. You should check this by expansion (multiplication) of the right-hand side.

Now apply this formula to the problem at hand by substituting $x = a$ and $y = 4$.

The difference of two cubes can also be factored, and the formula for the sum of two cubes can be used to do it:

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

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

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