Answer
The solution is $x = \frac{4}{3}, \frac{-2 - 2i\sqrt {3}}{3}, \text{ and } \frac{-2 + 2i\sqrt {3}}{3}$.
Work Step by Step
Factor out $3$:
$3(27x^3 - 64) = 0$
Divide both sides by $3$ to get:
$27x^3 - 64 = 0$
We see that $27x^3 - 64 = (3x)^3-4^3$, which is a difference 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 $a = \sqrt[3] {27x^3}$ (or $a = 3x$) and $b = \sqrt[3] {64}$ (or $b = 4$:):
$(3x - 4)((3x)^2 + (3x)(4) + 4^2) = 0$
$(3x - 4)(9x^2 + 12x + 16) = 0$
Use the Zero-Product Property by equating each factor to $0$, then solve each equation:
First factor:
$3x - 4 = 0$
$3x = 4$
$x = \frac{4}{3}$
Second factor:
$9x^2 + 12x + 16 = 0$
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 $a=9, b=12,$ and $c=16$ is the constant.
Let us plug in the numbers into the formula:
$x = \dfrac{-12 \pm \sqrt {(12)^2 - 4(9)(16)}}{2(9)}$
$x = \dfrac{-12 \pm \sqrt {144 - 576}}{18}$
$x = \dfrac{-12 \pm \sqrt {-432}}{18}$
The number $-432$ can be expanded into the factors $-144$ and $3$:
$x = \dfrac{-12 \pm \sqrt {(-144)(3)}}{18}$
We can remove the $-144$ from under the radical because the square root of $-144$ is $12i$:
$x = \dfrac{-12 \pm 12i\sqrt {3}}{18}$
Divide both numerator and denominator by $6$ to reduce the fraction:
$x = \dfrac{-2 \pm 2i\sqrt {3}}{3}$
The solution is $x = \dfrac{4}{3}, \dfrac{-2 \pm 2i\sqrt {3}}{3}$.
