Question

If we have the cube roots of unity as $1,\omega ,{{\omega }^{2}}$ then prove that $\left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)=49$

Answer 1

Hint: To find the value of the $\left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)$ , we need to derive ${{\omega }^{10}}$ and ${{\omega }^{11}}$ in the terms $\left( z-{{\omega }^{10}} \right)$ and $\left( z-{{\omega }^{11}} \right)$ and then simplify the left hand side.

Complete step-by-step solution -
Here, we are given that the cube roots of unity as $1,\omega ,{{\omega }^{2}}$ . The equation is given as:
${{x}^{3}}-1=0..............\left( i \right)$
As given in question $1,\omega$ and ${{\omega }^{2}}$ are the roots of this equation. Now, we will put $\omega$ in place of $x$ as it is one of the root. After doing this, we will get following:
${{\omega }^{3}}-1-0................\left( ii \right)$
In the above equation, we are going to use identity ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$ . Thus, applying this identity in the above equation, we will get:
$\left( \omega -1 \right)\left( {{\omega }^{2}}+\omega +1 \right)=0$
In the above equation, we have got the choices either $\left( \omega -1 \right)=0$ or $\left( {{\omega }^{2}}+\omega +1 \right)=0$ . Let us consider two cases
Case 1:
$\begin{aligned} & \omega -1=0 \\\ & \Rightarrow \omega =1 \\\ \end{aligned}$
It is not possible since $\omega$ is an imaginary number.
Case 2: ${{\omega }^{2}}+\omega +1=0.............\left( iii \right)$
When we will solve this equation, we will get imaginary values of $\omega$ . Also ${{\omega }^{2}}+\omega +1=0$ is an important result which we are going to use in our solution.
Now from equation (ii), we have ${{\omega }^{3}}-1=0$
$\Rightarrow {{\omega }^{3}}=1................\left( iv \right)$
Now, we will multiply both sides of the equation with ${{\omega }^{3}}$ . After doing this we will get:

& {{\omega }^{3}}\times {{\omega }^{3}}={{\omega }^{3}} \\\ & {{\omega }^{6}}={{\omega }^{3}}.............\left( v \right) \\\ \end{aligned}$$ From equation (iv) and (v), we get: ${{\omega }^{6}}=1............\left( vi \right)$ Now, we will multiply with ${{\omega }^{3}}$ on both sides of the above equation. After doing this we will get: $\begin{aligned} & {{\omega }^{6}}\times {{\omega }^{3}}={{\omega }^{3}} \\\ & \Rightarrow {{\omega }^{9}}={{\omega }^{3}}............\left( vii \right) \\\ \end{aligned}$ From equation (iv) and (vii), we will get: ${{\omega }^{9}}=1.............\left( viii \right)$ Now, we will multiply both sides of the equation with $\omega $ . After doing this, we will get: ${{\omega }^{10}}=\omega .........\left( ix \right)$ We will multiply both sides of the equation (ix) with $\omega $ . After doing this we will get ${{\omega }^{11}}={{\omega }^{2}}...........\left( x \right)$ Now, we are going to put the values of ${{\omega }^{11}}$ and ${{\omega }^{10}}$ from equation (x) and (ix) into $\left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)$ . After doing this we will get: $$\begin{aligned} & \left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)=\left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right) \\\ & \Rightarrow \left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)={{\left( 2-\omega \right)}^{2}}{{\left( 2-{{\omega }^{2}} \right)}^{2}} \\\ & \Rightarrow \left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)={{\left[ \left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right) \right]}^{2}} \\\ & \Rightarrow \left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)={{\left[ 4-2{{\omega }^{2}}-2\omega +{{\omega }^{3}} \right]}^{2}} \\\ & \Rightarrow \left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)={{\left[ 4-2\left( \omega +{{\omega }^{2}} \right)+{{\omega }^{3}} \right]}^{2}}................\left( xi \right) \\\ \end{aligned}$$ We know that the values of ${{\omega }^{3}}=1$ . Also, we know from equation (iii) that, $\begin{aligned} & 1+\omega +{{\omega }^{2}}=0 \\\ & \Rightarrow {{\omega }^{2}}+\omega =-1 \\\ \end{aligned}$ Using above values in equation (xi), we get $$\begin{aligned} & \Rightarrow \left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)={{\left[ 4-2\left( -1 \right)+1 \right]}^{2}} \\\ & \Rightarrow \left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)={{\left[ 7 \right]}^{2}} \\\ & \Rightarrow \left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)=49 \\\ \end{aligned}$$ Hence proved Note: In the equation (xi), instead of using the formula $$1+\omega +{{\omega }^{2}}=0$$ , we could have used the actual values of $\omega $ and ${{\omega }^{2}}$. The value of $\omega =\dfrac{-1+\sqrt{3}i}{2}$ and the value of ${{\omega }^{2}}=\dfrac{-1-\sqrt{3}i}{2}$ .Also, we could have solved the above question by expanding the term $$\left( 2-\omega \right)\left( 2-{{\omega }^{2}} \right)\left( 2-{{\omega }^{10}} \right)\left( 2-{{\omega }^{11}} \right)$$ . The answer would still have been same.