Q69 / 180 · COMEDK 2012 · Mathematics

If $1, \omega, \omega^{2}$ are the cube roots of unity, then $\left(3+3 \omega^{2}+5 \omega\right)^{6}-\left(2+6 \omega^{2}+2 \omega\right)^{3}$ is equal to

  • A
    32
  • B
    64
  • Correct answer
    C
    0
  • D
    $-1$

Solution

We have, $$ \begin{aligned} (3&\left.+3 \omega^{2}+5 \omega\right)^{6}-\left(2+6 \omega^{2}+2 \omega\right)^{3} \\ &=\left[3\left(1+\omega^{2}\right)+5 \omega\right]^{6}-\left[2(1+\omega)+6 \omega^{2}\right]^{3} \\ &=[3(-\omega)+5 \omega]^{6}-\left[2\left(-\omega^{2}\right)+6 \omega^{2}\right]^{3} \\ &=(2 \omega)^{6}-\left(4 \omega^{2}\right)^{3} \\ &=2^{6} \omega^{6}-64 \omega^{6} \\ &=64-64=0 \end{aligned} $$

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

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