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If $\omega$ denotes the cube root of unity, then what is the real root of the equation $x^{3}-27=0$ ?
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The correct answer is:
$3 \omega^{3}$
The given equation is $x^{3}-27=0$ $\Rightarrow \mathrm{x}^{3}-3^{3}=0 \Rightarrow \mathrm{x}=3,3 \omega, 3 \omega^{2}$
Thus, real root of given equation is $3 \omega^{3}$, since $\omega^{3}=1$
Thus, real root of given equation is $3 \omega^{3}$, since $\omega^{3}=1$
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