Answer
The solution is $x = \dfrac{2}{3}, \dfrac{-1 - i\sqrt {3}}{3}, \text{ and } \dfrac{-1 + i\sqrt {3}}{3}$.
Work Step by Step
Rewrite the equation so that all terms are on the left side of the equation:
$27x^3 - 8 = 0$
We see that $27x^3 - 8=(3x)^3-2^3$ 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 $b = \sqrt[3] {8}$ (or $b = 2$):
$(3x - 2)((3x)^2 + 3x(2) + 2^2) = 0$
$(3x - 2)(9x^2 + 6x + 4) = 0$
The zero product property states that if the product of two factors equals zero, then either one of the factors is zero or both factors equal zero. We can, therefore, set each factor to $0$:
Let us look at the first factor:
$3x - 2 = 0$
$3x = 2$
Divide each side by $3$ to solve for $x$:
$x = \frac{2}{3}$
We look at the other factor:
$9x^2 + 6x + 4 = 0$
We cannot factor this polynomial, so we resort to using the quadratic formula, which is $x = \dfrac{-b ± \sqrt {b^2 - 4ac}}{2a}$ where $a=9, b=6,$ and $c=4$.
$x = \dfrac{-6 \pm \sqrt {(6)^2 - 4(9)(4)}}{2(9)}$
$x = \dfrac{-6 \pm \sqrt {36 - 144}}{18}$
$x = \dfrac{-6 \pm \sqrt {-108}}{18}$
The number $-108$ can be expanded into the factors $-36$ and $3$:
$x = \dfrac{-6 \pm \sqrt {(-36)(3)}}{18}$
We can take out $-36$ from the radical because the square root of $-36$ is $6i$:
$x = \dfrac{-6 \pm 6i\sqrt {3}}{18}$
Divide all terms by $6$ to simplify the fraction:
$x = \frac{-1 \pm i\sqrt {3}}{3}$
The solutions are $x = \dfrac{2}{3}, \dfrac{-1 ± i\sqrt {3}}{3}$.
