Answer
The solutions are $x = \dfrac{3}{5}, \dfrac{-3 - 3i\sqrt {3}}{10}, \text{ and } \dfrac{-3 - 3i\sqrt {3}}{10}$.
Work Step by Step
We see that $125x^3 - 27$ 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 $a = \sqrt[3] {125x^3}$ (or $a = 5x$) and $b = \sqrt[3] {27}$ (or $b = 3$:):
$(5x - 3)((5x)^2 + 5x(3) + 3^2) = 0$
$(5x - 3)(25x^2 + 15x + 9) = 0$
We equate each factor to $0$ then solve each equation.
First factor:
$5x - 3 = 0$
$5x = 3$
$x = \dfrac{3}{5}$
Second factor:
$25x^2 + 15x + 9 = 0$
We cannot factor this polynomial, so we resort to using the quadratic formula, which is:
$x = \dfrac{-b ± \sqrt {b^2 - 4ac}}{2a}$
where $a$ is the coefficient of the first term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant.
Substitute $a=25, b=15, \text{ and } c=9$ into the formula:
$x = \dfrac{-15) ± \sqrt {(15)^2 - 4(25)(9)}}{2(25)}$
$x = \dfrac{-15) ± \sqrt {225 - 900}}{50}$
$x = \dfrac{-15) ± \sqrt {-675}}{50}$
The number $-675$ can be expanded into the factors $-225$ and $3$:
$x = \dfrac{-15) ± \sqrt {(-225)(3)}}{50}$
We can take out $-225$ from the radical because the square root of $-225$ is $15i$:
$x = \dfrac{-15 ± 15i\sqrt {3}}{50}$
$x = \dfrac{-3 ± 3i\sqrt {3}}{10}$
The solutions are $x = \dfrac{3}{5}, \dfrac{-3 - 3i\sqrt {3}}{10}, \text{ and } \dfrac{-3 - 3i\sqrt {3}}{10}$.
