Let $f:C\to C$ be a complex-valued function defined by $f\left( x \right)={{x}^{3}}$. Determine the set ${{f}^{-1}}\left( -1 \right)$

Hint: Use the definition of ${{f}^{-1}}\left( -1 \right)$ as $\left\{ x:{{x}^{3}}=1,x\in C \right\}$.Use the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and zero product property. Use the quadratic formula for solving the quadratic equation generated while solving the question.

Complete step-by-step answer:

Let $x\in {{f}^{-1}}\left( -1 \right)$
Since ${{f}^{-1}}\left( -1 \right)=\left\{ x:{{x}^{3}}=1,x\in C \right\}$ we have ${{x}^{3}}=-1$
Adding 1 on both sides, we get
${{x}^{3}}+1=0$
We know that ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$
Using the above formula, we get
$\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)=0$
From zero product property, we know that if ab = 0 then, a= 0 or b = 0.
Hence we have
$x+1=0$ or ${{x}^{2}}-x+1=0$
If x + 1 = 0, then x = -1
If ${{x}^{2}}-x+1=0$ then we know that roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here a = 1, b = -1 , c= 1
Using the above formula, we get
$x=\dfrac{1\pm \sqrt{1-4}}{2}=\dfrac{1\pm \sqrt{-3}}{2}=\dfrac{1\pm i\sqrt{3}}{2}$
Taking + sign we get $x=\dfrac{1+i\sqrt{3}}{2}$
Taking – sign we get $x=\dfrac{1-i\sqrt{3}}{2}$
Hence we have ${{f}^{-1}}\left( -1 \right)=\left\{ -1,\dfrac{1+i\sqrt{3}}{2},\dfrac{1-i\sqrt{3}}{2} \right\}$

Note: Alternate solution 1.
We know that if ${{x}^{3}}=1$ then $x=1,\omega ,{{\omega }^{2}}$ where $\omega =\dfrac{-1+i\sqrt{3}}{2}$
Now we have ${{x}^{3}}=-1$
Put t = -x
So, we have ${{t}^{3}}=1$
Hence $t=1,\omega ,{{\omega }^{2}}$
Reverting to the original variable, we get
$\begin{align}
  & -x=1,\omega ,{{\omega }^{2}} \\
 & \Rightarrow x=-1,-\omega .-{{\omega }^{2}} \\
 & \Rightarrow x=-1,\dfrac{1-i\sqrt{3}}{2},\dfrac{1+i\sqrt{3}}{2} \\
\end{align}$
which is the same as obtained above.
Alternate Solution 2.
We know that ${{e}^{i\left( 2k+1 \right)\pi }}=-1,k\in \mathbb{Z}$
Hence we have
\[\begin{align}
  & {{x}^{3}}={{e}^{i\left( 2k+1 \right)\pi }} \\
 & \Rightarrow x={{e}^{i\dfrac{\left( 2k+1 \right)\pi }{3}}},k=0,1,2 \\
\end{align}\]
Hence we have \[x={{e}^{\dfrac{\pi }{3}i}},{{e}^{\dfrac{3\pi }{3}i}},{{e}^{\dfrac{5\pi }{3}i}}\]
Using ${{e}^{ix}}=\cos x+i\sin x$, we get
$x=\dfrac{1+i\sqrt{3}}{2},-1,\dfrac{1-i\sqrt{3}}{2}$ which is the same as obtained above
Alternate Solution 3:
If $1,\alpha ,{{\alpha }^{2}},\ldots ,{{\alpha }^{n-1}}$ are the nth roots of unity
then the solutions of the equation ${{x}^{n}}=a,a\in R$ , such that there exists at least one real root, are $p,p\alpha ,p{{\alpha }^{2}},\ldots ,p{{\alpha }^{n-1}}$ where p is one of the real roots of the equations,e.g. $p={{a}^{\dfrac{1}{n}}}$
Taking n = 3 , a = -1 and $\alpha =\omega $ we get the desired result.