# How do you factor 64x^3-1?

Jan 16, 2016

$64 {x}^{3} - 1 = \left(4 x - 1\right) \left(16 {x}^{2} + 4 x + 1\right)$

$= \left(4 x - 1\right) \left(4 x + \frac{1}{2} - \frac{\sqrt{3}}{2} i\right) \left(4 x + \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$

#### Explanation:

The difference of cubes formula states that:
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
(try multiplying the right side to verify this)

Applying this, we have:

$64 {x}^{3} - 1 = {\left(4 x\right)}^{3} - {1}^{3}$
$= \left(4 x - 1\right) \left({\left(4 x\right)}^{2} + 4 x \cdot 1 + {1}^{2}\right)$
$= \left(4 x - 1\right) \left(16 {x}^{2} + 4 x + 1\right)$

If we wished to factor further, we could apply the quadratic formula to find the roots of $16 {x}^{2} + 4 x + 1$.

$16 {x}^{2} + 4 x + 1 = 0 \implies x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(16\right) \left(1\right)}}{2 \cdot 16}$

$\implies x = \frac{- 4 \pm \sqrt{- 48}}{32}$

$\implies x = \frac{- 1 \pm \sqrt{- 3}}{8}$

As we have the square root of a negative number, we have no real solutions, and so are done if we are using only real numbers. If we allow for complex numbers, then using $i = \sqrt{- 1}$, we have:

$x = - \frac{1}{8} \pm \frac{\sqrt{3}}{8} i$

and thus, remembering to multiply by $16$ to account for the coefficient of ${x}^{2}$, we can completely factor the original expression as

$64 {x}^{3} - 1 = \left(4 x - 1\right) \left(4 x + \frac{1}{2} - \frac{\sqrt{3}}{2} i\right) \left(4 x + \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$