Question

Prove the following equation where $\omega $ is the imaginary cube root of unity.
 $\dfrac{1}{{1 + 2\omega }} + \dfrac{1}{{2 + \omega }} - \dfrac{1}{{1 + \omega }} = 0$

Answer 1

Hint- In this question we should know that the imaginary cube roots of unity are given as 1, $\omega $ and ${\omega ^2}$. Also one should know that since $\omega^3 = 1$. And using the property: The sum of three cube roots of unity is zero i.e., $1 + \omega + \omega^2 = 0$, we will get the solution.

Complete step-by-step answer:
By taking LHS,
$
  LHS = \dfrac{1}{{1 + 2\omega }} + \dfrac{1}{{2 + \omega }} - \dfrac{1}{{1 - \omega }} \\
  \Rightarrow LHS = \dfrac{1}{{1 + 2\omega }} + \dfrac{{1 + \omega - 2 - \omega }}{{\left( {2 + \omega } \right)\left( {1 + \omega } \right)}} \\
  \Rightarrow LHS = \dfrac{1}{{1 + 2\omega }} + \dfrac{{ - 1}}{{2 + 2\omega + \omega + {\omega ^2}}} \\
  \Rightarrow LHS = \dfrac{1}{{1 + 2\omega }} + \dfrac{{ - 1}}{{1 + 2\omega + 1 + \omega + {\omega ^2}}} \\
 $
As($\omega $ denotes omega/cube root of unity )
so $1 + \omega + {\omega ^2} = 0$
$
  \Rightarrow LHS = \dfrac{1}{{1 + 2\omega }} - \dfrac{1}{{1 + 2\omega }} \\
  \Rightarrow LHS = \dfrac{{1 - 1}}{{1 + 2\omega }} \\
  \Rightarrow LHS = 0 \\
  \Rightarrow LHS = RHS \\
 $
Hence proved.

Note- Whenever we come up with this type of problem, one should know that cube roots of unity can be defined as the numbers which when raised to the power of 3 gives the result as 1. In simple words, the cube root of unity is the cube root of 1 that is $\sqrt[3]{1}$. By using the properties of cube roots of unity one can easily solve these types of questions.