Answer
The solutions are $\dfrac{1}{4}, \dfrac{-1 - i\sqrt {3}}{8}, \text{ and } \dfrac{-1 + i\sqrt {3}}{8}$.
Work Step by Step
We see that $64x^3 - 1$ is the difference of two cubes. We can factor using the formula:
$(a - b)(a^2 + ab + b^2)$
We plug in the values, where $a = \sqrt[3]{64x^3}$ (or $a = 4x$) and $b = \sqrt[3]{1}$ (or $b = 1$:):
$(4x - 1)((4x)^2 + 4x(1) + 1^2) = 0$
$(4x - 1)(16x^2 + 4x + 1) = 0$
Equate each factor to $0$, then solve each equation.
First factor:
$4x - 1 = 0$
$4x = 1$
$x = \frac{1}{4}$
Second factor:
$16x^2 + 4x + 1 = 0$
We cannot factor this polynomial, so we resort to using the quadratic formula, which is:
$x = \frac{-b ± \sqrt {b^2 - 4ac}}{2a}$
where $a$ is the coefficient of the first term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant.
Substitute $a=16, b=4, \text{ and } c=1$ into the formula:
$x = \dfrac{-4 \pm \sqrt {(4)^2 - 4(16)(1)}}{2(16)}$
$x = \dfrac{-4 \pm \sqrt {16 - 64}}{32}$
$x = \dfrac{-4 \pm \sqrt {-48}}{32}$
The number $-48$ can be expanded into the factors $-16$ and $3$:
$x = \dfrac{-4 ± \sqrt {(-16)(3)}}{32}$
We can take out $-16$ from the radical because the square root of $-16$ is $4i$:
$x = \dfrac{-4 \pm 4i\sqrt {3}}{32}$
Divide all terms by $4$ to simplify the fraction:
$x = \dfrac{-1 \pm i\sqrt {3}}{8}$
The solutions are $\dfrac{1}{4}, \dfrac{-1 - i\sqrt {3}}{8}, \text{ and } \dfrac{-1 + i\sqrt {3}}{8}$.
