Q93 / 180 · COMEDK 2013 · Mathematics

If $\omega=\frac{-1+\sqrt{3 i}}{2} \operatorname{then}\left(3+\omega+3 \omega^{2}\right)^{4}$ is

  • A
    $16$
  • B
    $-16$
  • Correct answer
    C
    $16 \omega$
  • D
    $16 \omega^{2}$

Solution

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 $\left.\omega^{3}=1\right]$ $=(-2 \omega)^{4}=16 \omega^{4}$ $=16 \omega \quad\left[\because \omega^{4}=\omega^{3} \cdot \omega=\omega\right]$

Hence, the correct answer is option: 'C'.

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