Question

If $\omega $is the complex cube root of unity, then the value of $\dfrac{{a + b\omega + c{\omega ^2}}}{{c + a\omega + b{\omega ^2}}} + \dfrac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}}$ is-
a)0
b)-1
c)1
d)2

Answer 1

Hint- we will take both the parts of this addition equation and solve them separately. After solving then separately and simplifying them to some extent, we will take both the simplified equations and add them in the end.

Complete step-by-step answer:

Breaking the given equation into two different equations, we have-

$\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}}}$ (given equation)

After breaking we will have two equations-

$\dfrac{{a + b\omega + c{\omega ^2}}}{{c + a\omega + b{\omega ^2}}}$ $ \to $

equation 1

$\dfrac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}}$ $ \to $

equation 2

Solving equation 1 first, we have-

$\dfrac{{a + b\omega + c{\omega ^2}}}{{c + a\omega + b{\omega ^2}}}$ (this is

equation 1)

Now, we will multiply both numerator and denominator by ${\omega ^2}$, we will get:

$

  \dfrac{{{\omega ^2}\left( {a + b\omega + c{\omega ^2}} \right)}}{{{\omega ^2}\left( {c +

 a\omega + b{\omega ^2}} \right)}} \\

    \\

   \Rightarrow \dfrac{{{\omega ^2}\left( {a + b\omega + c{\omega ^2}} \right)}}{{\left(

{c{\omega ^2} + a{\omega ^3} + b{\omega ^4}} \right)}} \\

$

Solving equation 2 now, we have-

$\dfrac{{a + b\omega + c{\omega ^2}}}{{b + c\omega + a{\omega ^2}}}$ (this is equation

 2)

Now, we will multiply both numerator and denominator by $\omega $, we will get:

$

  \dfrac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\omega \left( {b + c\omega +

a{\omega ^2}} \right)}} \\

    \\

   \Rightarrow \dfrac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {b\omega

+ c{\omega ^2} + a{\omega ^3}} \right)}} \\

$

Now, let $\dfrac{{{\omega ^2}\left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {c{\omega

^2} + a{\omega ^3} + b{\omega ^4}} \right)}}$ be equation 3 and $\dfrac{{\omega \left( {a +

b\omega + c{\omega ^2}} \right)}}{{\left( {b\omega + c{\omega ^2} + a{\omega ^3}}

\right)}}$ be equation 4, we get-

$\dfrac{{{\omega ^2}\left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {c{\omega ^2} +

a{\omega ^3} + b{\omega ^4}} \right)}}$ $ \to $ equation 3

$\dfrac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {b\omega + c{\omega

^2} + a{\omega ^3}} \right)}}$ $ \to $ equation 4

Adding the equations mentioned above i.e. equation number 3 and equation number 4, we

will have this equation-

$\dfrac{{{\omega ^2}\left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {c{\omega ^2} +

a{\omega ^3} + b{\omega ^4}} \right)}} + \dfrac{{\omega \left( {a + b\omega + c{\omega ^2}}

\right)}}{{\left( {b\omega + c{\omega ^2} + a{\omega ^3}} \right)}}$

Solving the equation further and taking the numerator common and then taking the LCM,

we will have:

$\dfrac{{{\omega ^2}\left( {a + b\omega + c{\omega ^2}} \right)}}{{\left( {a + b\omega +

c{\omega ^2}} \right)}} + \dfrac{{\omega \left( {a + b\omega + c{\omega ^2}} \right)}}{{\left(

{a + b\omega + c{\omega ^2}} \right)}}$

Cancelling the equation $a + b\omega + c{\omega ^2}$ which is same in both the equations

in numerator and denominator-

${\omega ^2} + \omega = - 1$

Hence the simplified value of the given equation is -1.

Note: To simplify the equations, always opt for the way of dividing them into two parts so that it won’t be congested and also solve them separately by naming the equations one or two. After solving separately, bring them together and carry on with the question.