Question

If $\alpha ,\beta $are the complex cube roots of unity, then ${\alpha ^4} + {\beta ^4} + {\alpha ^{ - 1}}{\beta ^{ - 1}} = $
$
  a.{\text{ 1}} \\
  {\text{b}}{\text{. }}\omega \\
  {\text{c}}{\text{. }}{\omega ^2} \\
  {\text{d}}{\text{. 0}} \\
$

Answer 1

Hint: - Use $\alpha = \omega ,{\text{ }}\beta = {\omega ^2}$

As we know if $\alpha $and $\beta $are the complex cube roots of unity therefore
$1 + \alpha + \beta = 0................\left( 1 \right)$
As we know cube roots of unity are $1,\omega ,{\omega ^2}$
Where $\omega $and${\omega ^2}$are non-real complex cube roots of unity therefore
${\omega ^3} = 1........\left( 2 \right),{\text{ }}1 + \omega + {\omega ^2} = 0...............\left( 3 \right)$
So, from equations (1) and (3)
$\alpha = \omega ,{\text{ }}\beta = {\omega ^2}$
Now given equation is ${\alpha ^4} + {\beta ^4} + {\alpha ^{ - 1}}{\beta ^{ - 1}}$
\[
   \Rightarrow {\omega ^4} + {\left( {{\omega ^2}} \right)^4} + {\omega ^{ - 1}}{\left( {{\omega ^2}} \right)^{ - 1}} \\
   \Rightarrow {\omega ^3}.\omega + {\omega ^8} + {\omega ^{ - 1}}\left( {{\omega ^{ - 2}}} \right) \\
   \Rightarrow {\omega ^3}.\omega + {\left( {{\omega ^3}} \right)^2}{\omega ^2} + \dfrac{1}{{{\omega ^3}}} \\
\]
From equation (2)
$
  {\omega ^3} = 1 \\
   \Rightarrow 1.\omega + {\left( 1 \right)^2}{\omega ^2} + 1 \\
   \Rightarrow 1 + \omega + {\omega ^2} \\
$
From equation (3)
$
  1 + \omega + {\omega ^2} = 0 \\
   \Rightarrow {\alpha ^4} + {\beta ^4} + {\alpha ^{ - 1}}{\beta ^{ - 1}} = 1 + \omega + {\omega ^2} = 0 \\
$
Hence, option (d) is correct.
Note: - Whenever we face such types of problems the key concept is that always remember the condition of cube roots of unity which is stated above, then substitute the values in the given equation then simplify we will get the required answer.