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Question: If \(1,w,{{w}^{2}}\) are three cube roots of unity, then \(\left( 1-w+{{w}^{2}} \right)\left( 1+w-{{...

If 1,w,w21,w,{{w}^{2}} are three cube roots of unity, then (1−w+w2)(1+w−w2)\left( 1-w+{{w}^{2}} \right)\left( 1+w-{{w}^{2}} \right) is:
(A) 1
(B) 2
(C) 3
(D) 4

Explanation

Solution

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 (a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+ce+cf\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 w3=1{{w}^{3}}=1 and 1+w+w2=01+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,
(1−w+w2)(1+w−w2)………(i)\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, (a+b+c)(d+e+f)=ad+ae+af+bd+be+bf+cd+ce+cf\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,
1(1)+1(w)+1(−w2)+(−w)(1)+(−w)(w)+(−w)(−w2)+w2(w)+w2(−w2) ⇒1+w−w2−w−w2+w3+w2+w3−w4 \begin{aligned} & 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{aligned}
As we know, the same terms with opposite signs cancel out each other, thus we get,
1+2w3−w2−w41+2{{w}^{3}}-{{w}^{2}}-{{w}^{4}}
Now, we know that w3=1{{w}^{3}}=1, therefore, we get,
1+2(1)−w2−w4 ⇒1+2−w2−w4 ⇒3−w2−w4 \begin{aligned} & 1+2\left( 1 \right)-{{w}^{2}}-{{w}^{4}} \\\ & \Rightarrow 1+2-{{w}^{2}}-{{w}^{4}} \\\ & \Rightarrow 3-{{w}^{2}}-{{w}^{4}} \\\ \end{aligned}
Now, we will write w4{{w}^{4}} as the product of ww and w3{{w}^{3}}. So, we get,
3−w2−w3.w3-{{w}^{2}}-{{w}^{3}}.w
So, again we will apply the formula, w3=1{{w}^{3}}=1, and we get,
3−w2−w3.w ⇒3−w2−w \begin{aligned} & 3-{{w}^{2}}-{{w}^{3}}.w \\\ & \Rightarrow 3-{{w}^{2}}-w \\\ \end{aligned}
Thus we know that 1+w+w2=01+w+{{w}^{2}}=0, so we will apply this in the above formula and we get,
3−(w2+w) ⇒3−[−1] \begin{aligned} & 3-\left( {{w}^{2}}+w \right) \\\ & \Rightarrow 3-\left[ -1 \right] \\\ \end{aligned}
On opening the brackets of all the above equation, we get,
3+1 ⇒4 \begin{aligned} & 3+1 \\\ & \Rightarrow 4 \\\ \end{aligned}

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.