# How do you solve x^3 + 64 = 0?

Jan 21, 2016

$x = - 4 , 2 \pm 2 \sqrt{3} i$

#### Explanation:

Notice that this is a sum of cubes, which is factorable as follows:

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

Thus, ${x}^{3} + 64$ is factorable into

${x}^{3} + {4}^{3} = \left(x + 4\right) \left({x}^{2} - 4 x + 16\right) = 0$

Now, we have one linear factor and one quadratic factor.

$\left(x + 4\right) \left({x}^{2} - 4 x + 16\right) = 0$

We can set each of these equal to $0$ individually to find the values of $x$ that make the whole expression equal $0$.

$x + 4 = 0 \text{ "=>" } x = - 4$

The next requires the quadratic formula.

${x}^{2} - 4 x + 16 = 0 \text{ "=>" } x = \frac{4 \pm \sqrt{16 - 64}}{2}$

$\implies x = \frac{4 \pm 4 \sqrt{3} i}{2} \text{ "=>" } x = 2 \pm 2 \sqrt{3} i$

These are two imaginary solutions.