Answer
The solutions are $ -4, 2 + 2i\sqrt {3}, \text{ and } 2 - 2i\sqrt {3}$.
Work Step by Step
We see that $x^3 + 64 = 0$ is the sum of two cubes. We can factor using the formula:
$(a + b)(a^2 - ab + b^2)$
We plug in the values, where $a = \sqrt[3]{x^3}$ (or $a = x$) and $b = \sqrt[3] 64$ (or $b = 4$:):
$(x + 4)(x^2 - 4x + 16) = 0$
We set each factor to $0$: then solve each equation.
First factor:
$x + 4 = 0$
$x = -4$
We look at the other factor:
$x^2 - 4x + 16 = 0$
We cannot factor this polynomial, so we resort to using the quadratic formula, which is:
$x = \dfrac{-b ± \sqrt {b^2 - 4ac}}{2a}$
where $a$ is the coefficient of the first term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant.
Substitute $a=1, b=-4$, and $c=16$ into the formula then simplify:
$x = \dfrac{-(-4) ± \sqrt {(-4)^2 - 4(1)(16)}}{2(1)}$
$x = \dfrac{4 ± \sqrt {16 - 64}}{2}$
$x = \dfrac{4 ± \sqrt {-48}}{2}$
$x = \dfrac{4 ± \sqrt {3(-16)}}{2}$
We can take out $-16$ from the radical because the square root of $-16$ is $4i$:
$x = \dfrac{4 ± 4i\sqrt {3}}{2}$
$x = 2 ± 2i\sqrt {3}$
Thus, the solutions are $ -4, 2 + 2i\sqrt {3}, \text{ and } 2 - 2i\sqrt {3}$.
