Question

Find the complex cube root of $8$.

Answer 1

Step 1: Generate algebraic expressions for the roots

The cube roots of $8$ can be algebraically expressed as,

$x^3-8=0.$

By factoring it we get

$\begin{array}{rcl}{x}^3-8& =& x^3-2^3\\ & =& (x-2)(x^2+2x+4)\left[\because {\mathbf{a}}^{\mathbf{3}}\mathbf{-}{\mathbf{b}}^{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}\mathbf{ab}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{)}\right]\end{array}$

Step 2: Solve for the factors

From the above equation,

$$\begin{gathered} \Rightarrow x-2=0 \\ \Rightarrow x=2 \end{gathered}$$

So, $2$ is one of the complex cube roots of $8$ (since all real numbers are a subset of complex numbers).

Also, consider the quadratic factor

$x^2+2x+4=0$

It can be solved using the quadratic formula,

$\begin{array}{rcl}\mathit{x}& \mathbf{=}& \frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}\end{array}$

Here,

$a=1$, $b=2$ and $c=4$

$\begin{array}{rcl}& \Rightarrow & x=\frac{-2\pm \sqrt{2^2-(4\times 1\times 4)}}{2\times 1}\\ & =& \frac{-2\pm \sqrt{-12}}{2}\\ & =& \frac{-2\pm \sqrt{-3\times 4}}{2}\\ & =& \frac{-2\pm 2\sqrt{-3}}{2}\\ & =& -1\pm \sqrt{3}i\end{array}$

Therefore, the complex cube roots of $8$ are $2$, $-1+\sqrt{3}i$ and $-1-\sqrt{3}i$.