Q110 / 160 · TS EAMCET 2010 · Mathematics

If $\omega$ is a complex cube root of unity, then $(x+1)(x+\omega)(x-\omega-1)$ is equal to

  • Correct answer
    A
    $x^3-1$
  • B
    $x^3+1$
  • C
    $x^3+2$
  • D
    $x^3-2$

Solution

$\omega \rightarrow$ cube root of unity ie, $\quad \omega^3=1$ and $1+\omega+\omega^2=0$ Then, $(x+1)(x+\omega)(x-\omega-1)$ $=(x+1)\left(x^2+\omega x-\omega x-\omega^2-x-\omega\right)$ $=(x+1)\left(x^2-x-\left(\omega+\omega^2\right)\right)$ $=(x+1)\left(x^2-x-(-1)\right)$ $\left(\because \omega+\omega^2=-1\right)$ $=(x+1)\left(x^2-x+1\right)$ $=(x+1)\left(x^2+1-x\right)$ $=\left(x^3-1\right)$

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

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