Question

If cube root of unity is $1,\omega ,{\omega ^2}$ then, find the value of $\dfrac{{(a + b\omega + c{\omega ^2})}}{{(c + a\omega + b{\omega ^2})}} + \dfrac{{(a + b\omega + c{\omega ^2})}}{{(b + c\omega + a{\omega ^2})}}$
A) $\omega $
B) $1$
C) $ - \omega $
D) $ - 1$

Answer 1

Hint: For this question consider the given expression, now multiply and divide by ${\omega ^2}$ for first term and $\omega $ for second term and solve using $\omega + {\omega ^2} = - 1$ and ${\omega ^3} = 1$ formula.

Complete step-by-step answer:
The given expression is,
$\Rightarrow \dfrac{{(a + b\omega + c{\omega ^2})}}{{(c + a\omega + b{\omega ^2})}} + \dfrac{{(a + b\omega + c{\omega ^2})}}{{(b + c\omega + a{\omega ^2})}}$
Now multiply and divide by ${\omega ^2}$ for first term and $\omega $ for second term,
$\Rightarrow \dfrac{{{\omega ^2}(a + b\omega + c{\omega ^2})}}{{(c{\omega ^2} + a{\omega ^3} + b{\omega ^4})}} + \dfrac{{\omega (a + b\omega + c{\omega ^2})}}{{(b\omega + c{\omega ^2} + a{\omega ^3})}}$---(1)
We know that, ${\omega ^3} = 1$
Applying this for equation (1) we get,
$ \Rightarrow \dfrac{{{\omega ^2}(a + b\omega + c{\omega ^2})}}{{(c{\omega ^2} + a + b{\omega ^4})}} + \dfrac{{\omega (a + b\omega + c{\omega ^2})}}{{(b\omega + c{\omega ^2} + a)}}$--(2)
Now we know, ${\omega ^3} = 1$
Multiply both sides by $\omega $ we get,
${\omega ^4} = \omega $---(3)
Substitute equation (3) in equation (2)
$ \Rightarrow \dfrac{{{\omega ^2}(a + b\omega + c{\omega ^2})}}{{(c{\omega ^2} + a + b\omega )}} + \dfrac{{\omega (a + b\omega + c{\omega ^2})}}{{(b\omega + c{\omega ^2} + a)}}$
Solving the above equation,
$ \Rightarrow \omega + {\omega ^2}$
We know that, $\omega + {\omega ^2} = - 1$
Then we get,
${\kern 1pt} \Rightarrow - 1$
Hence, the solution is $ - 1$.

Note: Whenever we face such types of problems the key concept is simply to make use of the cube root of unity. In this problem another method is taking LCM and solving which consumes a lot of steps. Here, using the cube root of unity method will help us arrive at the solution easily.