Answer
The solutions are $3, -\dfrac{3 - 3i\sqrt {3}}{2}, \text{ and }-\dfrac{3 + 3i\sqrt {3}}{2}$.
Work Step by Step
We see that $x^3 - 27$ 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 $a = \sqrt[3]{x^3}$ (or $a = x$) and $b = \sqrt[3]{27}$ (or $b = 3$:):
$(x - 3)(x^2 + 3x + 3^2) = 0$
$(x - 3)(x^2 + 3x + 9) = 0$
Use the Zero-Product Property by equating each factor to $0$, then solve each equation.
First factor:
$x - 3 = 0$
$x = 3$
Second factor:
$x^2 + 3x + 9 = 0$
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 $a$ is the coefficient of the first term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant.
Substitute $a=1, b=3, \text{ and } c=9$ into the formula:
$x = \dfrac{-3 \pm \sqrt {(3)^2 - 4(1)(9)}}{2(1)}$
$x = \dfrac{-3 \pm \sqrt {9 - 36}}{2}$
$x = \dfrac{-3 \pm\sqrt {-27}}{2}$
$x = \dfrac{-3 \pm \sqrt {(-9)(3)}}{2}$
We can take out $-9$ from the radical because the square root of $-9$ is $3i$:
$x = \dfrac{-3 \pm 3i\sqrt {3}}{2}$
The solutions are $3, -\dfrac{3 - 3i\sqrt {3}}{2}, \text{ and }-\dfrac{3 + 3i\sqrt {3}}{2}$.
