Answer
The solutions are $x = -\dfrac{1}{2}, \dfrac{1 - i \sqrt {3}}{4}, \text{ and } \dfrac{1 + i \sqrt {3}}{4}$.
Work Step by Step
Rewrite the equation so that all the terms are on the left side of the equation and that the $0$ is on the right side of the equation:
$64x^3 + 8 = 0$
We see that $64x^3 + 8 = 0=(4x)^3+2^3$ is a sum of two cubes. We can factor using the formula:
$a^3+b^3=(a + b)(a^2 - ab + b^2)$
We plug in the values, where $a = \sqrt[3] {64x^3}$ (or $a = 4x$) and $b = \sqrt[3] {8}$ (or $b = 2$):
$(4x + 2)((4x)^2 - 4x(2) + 2^2) = 0$
$(4x + 2)(16x^2 - 8x + 4) = 0$
Use the Zero Product Property by equating each factor to $0$, then solve each equation.
First factor:
$4x + 2 = 0$
$4x = -2$
$x = -\frac{2}{4}$
$x = -\frac{1}{2}$
We look at the other factor:
$16x^2 - 8x + 4 = 0$
We cannot factor this polynomial, so we resort to using the quadratic formula, which is:
$x = \dfrac{-b \pm \sqrt {b^2 - 4ac}}{2a}$
where $a=16, b=-8,$ and $c=4$.
Let us plug in the numbers into the formula:
$x = \dfrac{-(-8) \pm \sqrt {-(-8)^2 - 4(16)(4)}}{2(16)}$
$x = \dfrac{8 \pm \sqrt {64 - 256}}{32}$
$x = \dfrac{8 \pm \sqrt {-192}}{32}$
The number $-192$ can be expanded into the factors $-64$ and $3$:
$x = \dfrac{8 \pm \sqrt {(-16)(3)}}{32}$
We can take out $-64$ from the radical because the square root of $-64$ is $8i$:
$x = \dfrac{8 \pm 8i \sqrt {3}}{32}$
$x = \dfrac{1 \pm i \sqrt {3}}{4}$
The solutions are $x = -\dfrac{1}{2}, \dfrac{1 ± i \sqrt {3}}{4}$.
