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If $\alpha, \beta$ are non-real cube roots of 2 , then $\alpha^6+\beta^6$ equals
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Since, $\alpha$ and $\beta$ are non-real cube roots of 2 .
$$
\begin{aligned}
\therefore \quad \alpha & =2^{1 / 3} \omega \text { and } \beta=2^{1 / 3} \omega^2 \\
\text { Now, } \alpha^6+\beta^6 & =\left(2^{1 / 3} \omega\right)^6+\left(2^{1 / 3} \omega^2\right)^6 \\
& =2^2 \omega^6+2^2 \omega^{12} \\
& =4\left(\omega^3\right)^2+4\left(\omega^3\right)^4 \\
& =4+4=8 \quad \quad\left[\because \omega^3=1\right]
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad \alpha & =2^{1 / 3} \omega \text { and } \beta=2^{1 / 3} \omega^2 \\
\text { Now, } \alpha^6+\beta^6 & =\left(2^{1 / 3} \omega\right)^6+\left(2^{1 / 3} \omega^2\right)^6 \\
& =2^2 \omega^6+2^2 \omega^{12} \\
& =4\left(\omega^3\right)^2+4\left(\omega^3\right)^4 \\
& =4+4=8 \quad \quad\left[\because \omega^3=1\right]
\end{aligned}
$$
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