# How do you factor x^3+8?

Mar 14, 2018

#### Explanation:

An interesting fact:

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

In ${x}^{3} + 8$, ${a}^{3} = {x}^{3}$ and ${b}^{3} = 8$

Let's solve for $a$ and $b$.

$\implies {a}^{3} = {x}^{3}$

$\implies \sqrt[3]{{a}^{3}} = \sqrt[3]{{x}^{3}}$

$\implies a = x$

Now for $b$.

$\implies {b}^{3} = 8$

$\implies \sqrt[3]{{b}^{3}} = \sqrt[3]{8}$

$\implies b = 2$

Plug these values into our equation.

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

$\left(x + 2\right) \left({x}^{2} - 2 x + 4\right)$ This is our answer!

If you want to factor this further, we let ${x}^{2} - 2 x + 4 = 0$ and solve the equation.

${x}^{2} - 2 x + 4 = 0$ Use the quadratic formula:
$\frac{- b \pm \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 \left(a\right)}$

Here, $a = 1$, $b = - 2$, and $c = 4$

$x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(1\right) \left(4\right)}}{2 \left(1\right)}$

$x = \frac{2 \pm \sqrt{4 - 16}}{2}$

$x = \frac{2 \pm \sqrt{- 12}}{2}$

$x = \frac{2 \pm 2 i \sqrt{3}}{2}$

$x = 1 \pm i \sqrt{3}$

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

(x+2)(x-1-isqrt3))(x-1+isqrt3))

Therefore, our factored form, in this case, would be
(x+2)(x-1-isqrt3))(x-1+isqrt3))