Question

If $1,\omega $ and ${{\omega }^{2}}$ are the cube roots of unity, prove that \[{{\left( 1-\omega +{{\omega }^{2}} \right)}^{6}}+{{\left( 1+{{\omega }^{2}}-\omega \right)}^{6}}=128\].

Answer 1

Hint: Use the identity of cube roots of unity in complex numbers to simplify the expression. Sum of cube roots of unity is equal to zero. Also, when cube roots of unity are raised to the power 3, it gives 1.

Complete step-by-step answer:
Cube roots of unity are solutions of the equation, ${{x}^{3}}=1$. This equation has three roots: $1,\omega \text{ and }{{\omega }^{2}}$. Here, 1 is the real root and $\omega \text{ and }{{\omega }^{2}}$ are complex roots, that means, they are complex numbers. It is a particular case of ${{n}^{th}}$ roots of unity. The value of $\omega \text{ and }{{\omega }^{2}}$ are $\dfrac{1-\sqrt{3}i}{2}\text{ and }\dfrac{-1-\sqrt{3}i}{2}$ respectively. These roots are used in different branches and topics of math like number theory. It is also called the de Moivre system.

Now, let us come to the question. Therefore,

$L.H.S={{\left( 1-\omega +{{\omega }^{2}} \right)}^{6}}+{{\left( 1-{{\omega }^{2}}+\omega \right)}^{6}}$

We know that,

$\begin{align}

  & 1+\omega +{{\omega }^{2}}=0 \\

 & \therefore 1+\omega =-{{\omega }^{2}}\text{ and 1+}{{\omega }^{2}}=-\omega \\

\end{align}$

Using the above two relations, we get,

$\begin{align}

  & L.H.S={{\left( 1+{{\omega }^{2}}-\omega \right)}^{6}}+{{\left( 1+\omega -{{\omega }^{2}} \right)}^{6}} \\

 & \text{ }={{\left( -\omega -\omega \right)}^{6}}+{{\left( -{{\omega }^{2}}-{{\omega }^{2}} \right)}^{6}} \\

 & \text{ }={{\left( -2\omega \right)}^{6}}+{{\left( -2{{\omega }^{2}} \right)}^{6}} \\

 & \text{ }={{2}^{6}}{{\omega }^{6}}+{{2}^{6}}{{\omega }^{12}} \\

\end{align}$

Taking ${{2}^{6}}$ common in the above expression, we get,

$L.H.S={{2}^{6}}\left( {{\omega }^{6}}+{{\omega }^{12}} \right)$

We know that, ${{\omega }^{3}}=1$, so the above expression can be simplified as,

$\begin{align}

  & L.H.S={{2}^{6}}\left( {{\left( {{\omega }^{3}} \right)}^{2}}+{{\left( {{\omega }^{3}} \right)}^{4}} \right) \\

 & \text{ }={{2}^{6}}\times \left( {{1}^{2}}+{{1}^{4}} \right) \\

 & \text{ }={{2}^{6}}\times \left( 1+1 \right) \\

 & \text{ }={{2}^{6}}\times 2 \\

 & \text{ }={{2}^{7}} \\

 & \text{ }=128 \\

 & \text{ }=R.H.S \\

\end{align}$



Note: The important thing to note is the approach to solve this question. As you can see that we have used the identity of the sum of cube roots of unity. It will be very difficult for us to solve this question if we try to expand the 6th power of the term. It will take much time and the probability of making mistakes will be more.