Answer
The solutions are $x=\dfrac{1}{2}, \dfrac{-1 - i\sqrt {3}}{4}, \text{ and } \dfrac{-1 + i\sqrt {3}}{4}$.
Work Step by Step
Let us rewrite the equation so that all terms are on the left side of the equation:
$8x^3 - 1 = 0$
We see that $8x^3 - 1=(2x)^3-1^3$ 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 $a = \sqrt[3] {8x^3}$ (or $a = 2x$) and $b = \sqrt[3] {1}$ (or $b = 1$):
$(2x - 1)((2x)^2 + 2x(1) + 1^2) = 0$
$(2x - 1)(4x^2 + 2x + 1) = 0$
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:
$2x - 1 = 0$
$2x = 1$
$x = \frac{1}{2}$
We look at the other factor:
$4x^2 + 2x + 1 = 0$
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 $a=4, b=2,$ and $c=1$.
Let us plug in the numbers into the formula:
$x = \dfrac{-2 \pm \sqrt {2^2 - 4(4)(1)}}{2(4)}$
$x = \dfrac{-2 \pm \sqrt {4 - 16}}{8}$
$x = \dfrac{-2 \pm \sqrt {-12}}{8}$
$x = \dfrac{-2 \pm \sqrt {(-4)(3)}}{8}$
We can take out $-4$ from the radical because the square root of $-4$ is $2i$:
$x = \dfrac{-2 \pm 2i \sqrt {3}}{8}$
$x = \dfrac{-1 \pm i\sqrt {3}}{4}$
The solutions are $x = \dfrac{1}{2}, \dfrac{-1 \pm i\sqrt {3}}{4}$.
