Question

If the cube roots of unity are defined as 1, $\omega $ and ${{\omega }^{2}}$, then find the roots of the equation ${{\left( x-2 \right)}^{3}}+8=0$?
(a) 2, $2\omega $, $2{{\omega }^{2}}$,
(b) 0, $2\left( 1-\omega \right)$, $2\left( 1-{{\omega }^{2}} \right)$,
(c) 2, 0, $\omega $,
(d) -2, $-2\omega $, $-2{{\omega }^{2}}$.

Answer 1

Hint: We start solving the problem by converting the given equation ${{\left( x-2 \right)}^{3}}+8=0$ into the form that we have 1 on the right side of the equation. Once we convert into the form we required, we use the cube roots of the 1 to get the cube roots of the given equation. We then use all these roots to find the value of x which are the values of required roots.

Complete step by step answer:
According to the problem, the roots of unity are 1, $\omega $ and ${{\omega }^{2}}$. We need to find the roots of the equation ${{\left( x-2 \right)}^{3}}+8=0$.
We have ${{\left( x-2 \right)}^{3}}+8=0$.
$\Rightarrow {{\left( x-2 \right)}^{3}}=-8$.
$\Rightarrow \dfrac{{{\left( x-2 \right)}^{3}}}{-8}=1$.
$\Rightarrow \dfrac{{{\left( x-2 \right)}^{3}}}{{{\left( -2 \right)}^{3}}}=1$.
$\Rightarrow {{\left( \dfrac{x-2}{-2} \right)}^{3}}=1$.
$\Rightarrow {{\left( \dfrac{2-x}{2} \right)}^{3}}=1$.
$\Rightarrow \left( \dfrac{2-x}{2} \right)={{\left( 1 \right)}^{\dfrac{1}{3}}}$.
We have $\dfrac{2-x}{2}$ as cube roots of 1 and we know the cube roots of 1 are 1, $\omega $ and ${{\omega }^{2}}$.
So, we get roots of $\dfrac{2-x}{2}$ as 1, $\omega $ and ${{\omega }^{2}}$. Now, we find all the values of x.
$\Rightarrow \dfrac{2-x}{2}=1$.
$\Rightarrow 2-x=2$.
$\Rightarrow x=2-2$.
$\Rightarrow x=0$ ---(1).
Now, we have $\dfrac{2-x}{2}=\omega $.
$\Rightarrow 2-x=2\omega $.
$\Rightarrow 2-2\omega =x$.
$\Rightarrow x=2\left( 1-\omega \right)$ ---(2).
Now, we have $\dfrac{2-x}{2}={{\omega }^{2}}$.
$\Rightarrow 2-x=2{{\omega }^{2}}$.
$\Rightarrow 2-2{{\omega }^{2}}=x$.
$\Rightarrow x=2\left( 1-{{\omega }^{2}} \right)$ ---(3).
From equations (1), (2) and (3), we have found the roots of the equation ${{\left( x-2 \right)}^{3}}+8=0$ as 0, $2\left( 1-\omega \right)$ and $2\left( 1-{{\omega }^{2}} \right)$.
∴ The roots of the equation ${{\left( x-2 \right)}^{3}}+8=0$ are 0, $2\left( 1-\omega \right)$ and $2\left( 1-{{\omega }^{2}} \right)$.

So, the correct answer is “Option B”.

Note: We can solve the problem by finding the roots in standard procedure. We know that $\omega $ and ${{\omega }^{2}}$ are the complex roots of unity and these roots contain imaginary parts of complex numbers. We get the value of the roots in complex numbers and we compare the obtained roots with $\omega $ and ${{\omega }^{2}}$ to convert into the required form of roots.