Question

If $1,w,{{w}^{2}}$ are three cube roots of unity, then $\left( 1-w+{{w}^{2}} \right)\left( 1+w-{{w}^{2}} \right)$ is:
(A) 1
(B) 2
(C) 3
(D) 4

Answer 1

Hint: In this question, we have to find the value of an equation. Thus, we will use cube roots of unity and basic mathematical rules to get the solution. First, we will apply the distributive property $\left( a+b+c \right)\left( d+e+f \right)=ad+ae+af+bd+be+bf+cd+ce+cf$. Then, we will cancel out the same terms with opposite signs. Thus, we will apply the formula ${{w}^{3}}=1$ and $1+w+{{w}^{2}}=0$ in the equation. In the end, we will apply the necessary calculations to get the solution.

Complete step by step answer:
According to the problem, we have to find the value of an equation. Thus we will apply the distributive property and cube roots of unity to get the solution. The equation given to us is,
$\left( 1-w+{{w}^{2}} \right)\left( 1+w-{{w}^{2}} \right)\ldots \ldots \ldots \left( i \right)$
So, let us first apply the distributive property, $\left( a+b+c \right)\left( d+e+f \right)=ad+ae+af+bd+be+bf+cd+ce+cf$ in equation (i). So, we get,
$\begin{align}
  & 1\left( 1 \right)+1\left( w \right)+1\left( -{{w}^{2}} \right)+\left( -w \right)\left( 1 \right)+\left( -w \right)\left( w \right)+\left( -w \right)\left( -{{w}^{2}} \right)+{{w}^{2}}\left( w \right)+{{w}^{2}}\left( -{{w}^{2}} \right) \\
 & \Rightarrow 1+w-{{w}^{2}}-w-{{w}^{2}}+{{w}^{3}}+{{w}^{2}}+{{w}^{3}}-{{w}^{4}} \\
\end{align}$
As we know, the same terms with opposite signs cancel out each other, thus we get,
$1+2{{w}^{3}}-{{w}^{2}}-{{w}^{4}}$
Now, we know that ${{w}^{3}}=1$, therefore, we get,
$\begin{align}
  & 1+2\left( 1 \right)-{{w}^{2}}-{{w}^{4}} \\
 & \Rightarrow 1+2-{{w}^{2}}-{{w}^{4}} \\
 & \Rightarrow 3-{{w}^{2}}-{{w}^{4}} \\
\end{align}$
Now, we will write ${{w}^{4}}$ as the product of $w$ and ${{w}^{3}}$. So, we get,
$3-{{w}^{2}}-{{w}^{3}}.w$
So, again we will apply the formula, ${{w}^{3}}=1$, and we get,
$\begin{align}
  & 3-{{w}^{2}}-{{w}^{3}}.w \\
 & \Rightarrow 3-{{w}^{2}}-w \\
\end{align}$
Thus we know that $1+w+{{w}^{2}}=0$, so we will apply this in the above formula and we get,
$\begin{align}
  & 3-\left( {{w}^{2}}+w \right) \\
 & \Rightarrow 3-\left[ -1 \right] \\
\end{align}$
On opening the brackets of all the above equation, we get,
$\begin{align}
  & 3+1 \\
 & \Rightarrow 4 \\
\end{align}$

So, the correct answer is “Option D”.

Note: While solving this problem, do mention all the steps properly to avoid confusion. Do mention the cube roots of the unity formula properly to get an accurate answer.