Answer
The solutions are $x = -5, \dfrac{5 - 5i\sqrt {3}}{2}, \text{ and } \dfrac{5 + 5i\sqrt {3}}{2}$.
Work Step by Step
We see that $x^3 + 125 =x^3+5^3$ is a sum 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 $a = \sqrt[3] {x^3}$ (or $a = x$) and $b = \sqrt[3] {125}$ (or $b = 5$):
$(x + 5)(x^2 - 5x + 25) = 0$
Use the Zero-Product Property by equating each factor to zero, then solve each equation.
First factor:
$x + 5 = 0$
$x = -5$
Second factor:
$x^2 - 5x + 25 = 0$
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 $a=1, b=-5,$ and $c=25$.
$x = \dfrac{-(-5) \pm \sqrt {(-5)^2 - 4(1)(25)}}{2(1)}$
$x = \dfrac{5 \pm \sqrt {25 - 100}}{2}$
$x = \dfrac{5 \pm \sqrt {-75}}{2}$
The number $-75$ can be factored as $-25(3)$:
$x = \dfrac{5 \pm \sqrt {(-25)(3)}}{2}$
We can take out $-25$ from the radical because the square root of $-25$ is $5i$:
$x = \dfrac{5 \pm 5i\sqrt {3}}{2}$
The solutions are $x = -5, \dfrac{5 ± 5i\sqrt {3}}{2}$.
