Search any question & find its solution
Question:
Answered & Verified by Expert
If $\omega=\frac{-1+\sqrt{3 i}}{2} \operatorname{then}\left(3+\omega+3 \omega^{2}\right)^{4}$ is
Options:
Solution:
1653 Upvotes
Verified Answer
The correct answer is:
$16 \omega$
We have,
$\left(3+\omega+3 \omega^{2}\right)^{4}$
$=\left(3+3 \omega+3 \omega^{2}-2 \omega\right)^{4}$
$=\left[3\left(1+\omega+\omega^{2}\right)-2 \omega\right]^{4}$
$=(3 \times 0-2 \omega)^{4}$
$\left[\because 1+\omega+\omega^{2}=0\right.$ or $\left.\omega^{3}=1\right]$
$=(-2 \omega)^{4}=16 \omega^{4}$
$=16 \omega \quad\left[\because \omega^{4}=\omega^{3} \cdot \omega=\omega\right]$
$\left(3+\omega+3 \omega^{2}\right)^{4}$
$=\left(3+3 \omega+3 \omega^{2}-2 \omega\right)^{4}$
$=\left[3\left(1+\omega+\omega^{2}\right)-2 \omega\right]^{4}$
$=(3 \times 0-2 \omega)^{4}$
$\left[\because 1+\omega+\omega^{2}=0\right.$ or $\left.\omega^{3}=1\right]$
$=(-2 \omega)^{4}=16 \omega^{4}$
$=16 \omega \quad\left[\because \omega^{4}=\omega^{3} \cdot \omega=\omega\right]$
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.