Answer
$\left\{2, -1\pm i\sqrt3\right\}$
Work Step by Step
Let's factor out a $2$ from all terms:
$2(x^3 - 8) = 0$
Divide both sides by $2$ to get rid of it:
$x^3 - 8 = 0\\
x^3=2^3$
We see that $x^3 - 2^3 = 0$ is the 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 $a = \sqrt[3] {x^3}$ (or $a = x$) and $b = 2$ to obtain:
$x-2^3=0\\
(x - 2)(x^2 + 2x + 2^2) = 0$
Let's simplify a bit:
$(x - 2)(x^2 + 2x + 4) = 0$
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's look at the first factor:
$x - 2 = 0$
Add $2$ to each side to solve for $x$:
$x = 2$
We look at the other factor:
$x^2 + 2x + 4 = 0$
We cannot factor this polynomial, so we resort to using the Quadratic Formula, which is:
$x = \dfrac{-b ± \sqrt {b^2 - 4ac}}{2a}$
where $a$ is the coefficient of the first term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant.
Let's plug in the numbers into the formula:
$x = \dfrac{-2 ± \sqrt {2^2 - 4(1)(4)}}{2(1)}$
Let's simplify it:
$x = \dfrac{-2 ± \sqrt {4 - 16}}{2}$
Let's simplify what is inside the radical:
$x = \dfrac{-2 ± \sqrt {-12}}{2}$
The number $-12$ can be expanded into the factors $-4$ and $3$:
$x = \dfrac{-2 ± \sqrt {(-4)(3)}}{2}$
We can take out $-4$ from the radical because the square root of $-4$ is $2i$:
$x = \dfrac{-2 ± 2i \sqrt {3}}{2}$
Divide all terms by $2$ to simplify the fraction:
$x = -1 ± i\sqrt {3}$
The solution is $x = -1 ± i\sqrt {3}$.
Therefore the real and imaginary solutions are: $\left\{2, -1\pm i\sqrt3\right\}$.
