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(A) Given that $\alpha$ and $\beta$ are non-real cube roots of $2$.Since the cube roots of $2$ are $2^{1/3}, 2^{1/3}\omega, 2^{1/3}\omega^2$,where $\o

Vedclass · Accepted Answer

$8$

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(A) Given that $\alpha$ and $\beta$ are non-real cube roots of $2$.Since the cube roots of $2$ are $2^{1/3}, 2^{1/3}\omega, 2^{1/3}\omega^2$,where $\omega$ is the complex cube root of unity.The non-real cube roots are $\alpha = 2^{1/3}\omega$ and $\beta = 2^{1/3}\omega^2$.Now,$\alpha^6 + \beta^6 = (2^{1/3}\omega)^6 + (2^{1/3}\omega^2)^6$.$= 2^2 \omega^6 + 2^2 \omega^{12}$.$= 4(\omega^3)^2 + 4(\omega^3)^4$.Since $\omega^3 = 1$,we have $4(1)^2 + 4(1)^4 = 4 + 4 = 8$.

If $\alpha, \beta$ are non-real cube roots of $2$,then $\alpha^6 + \beta^6$ equals

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