Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If $\alpha$ and $\beta$ are the complex cube roots of unity, then $\alpha^3+\beta^3+\alpha^{-2} \times \beta^{-2}$ is equal to
MathematicsComplex NumberMHT CETMHT CET 2022 (10 Aug Shift 2)
Options:
  • A $1$
  • B $-3$
  • C $3$
  • D $0$
Solution:
2828 Upvotes Verified Answer
The correct answer is: $3$
$\begin{aligned} & \alpha^3+\beta^3+\alpha^{-2} \cdot \beta^{-2}=\alpha^3+\beta^3+\frac{1}{\alpha^2 \beta^2} \\ & =\omega^3+\omega^6+\frac{1}{\omega^2 \omega^4}=1+1+1=3\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.