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If $\omega$ is a complex cube root of unity, then $(x+1)(x+\omega)(x-\omega-1)$ is equal to
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Verified Answer
The correct answer is:
$x^3-1$
$\omega \rightarrow$ cube root of unity
ie, $\quad \omega^3=1$ and $1+\omega+\omega^2=0$
Then, $(x+1)(x+\omega)(x-\omega-1)$
$=(x+1)\left(x^2+\omega x-\omega x-\omega^2-x-\omega\right)$
$=(x+1)\left(x^2-x-\left(\omega+\omega^2\right)\right)$
$=(x+1)\left(x^2-x-(-1)\right)$
$\left(\because \omega+\omega^2=-1\right)$
$=(x+1)\left(x^2-x+1\right)$
$=(x+1)\left(x^2+1-x\right)$
$=\left(x^3-1\right)$
ie, $\quad \omega^3=1$ and $1+\omega+\omega^2=0$
Then, $(x+1)(x+\omega)(x-\omega-1)$
$=(x+1)\left(x^2+\omega x-\omega x-\omega^2-x-\omega\right)$
$=(x+1)\left(x^2-x-\left(\omega+\omega^2\right)\right)$
$=(x+1)\left(x^2-x-(-1)\right)$
$\left(\because \omega+\omega^2=-1\right)$
$=(x+1)\left(x^2-x+1\right)$
$=(x+1)\left(x^2+1-x\right)$
$=\left(x^3-1\right)$
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