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If $\alpha$ and $\beta$ are the complex cube roots of unity, then what is the value of $(1+\alpha)(1+\beta)\left(1+\alpha^{2}\right)\left(1+\beta^{2}\right) ?$
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Since, $\alpha$ and $\beta$ are the complex cube roots of unity therefore $1+\alpha+\alpha^{2}=0=1+\beta+\beta^{2}$
and $\alpha^{3}=1=\beta^{3}$.
Consider $(1+\alpha)(1+\beta)\left(1+\alpha^{2}\right)\left(1+\beta^{2}\right)$
$=(1+\alpha)\left(1+\alpha^{2}\right)(1+\beta)\left(1+\beta^{2}\right)$
$=\left(1+\alpha^{2}+\alpha+\alpha^{3}\right)\left(1+\beta^{2}+\beta+\beta^{3}\right)$
$=\left(0+\alpha^{3}\right)\left(0+\beta^{3}\right)$
$=\left(\alpha^{3}\right)\left(\beta^{3}\right)=(1)(1)=1$
and $\alpha^{3}=1=\beta^{3}$.
Consider $(1+\alpha)(1+\beta)\left(1+\alpha^{2}\right)\left(1+\beta^{2}\right)$
$=(1+\alpha)\left(1+\alpha^{2}\right)(1+\beta)\left(1+\beta^{2}\right)$
$=\left(1+\alpha^{2}+\alpha+\alpha^{3}\right)\left(1+\beta^{2}+\beta+\beta^{3}\right)$
$=\left(0+\alpha^{3}\right)\left(0+\beta^{3}\right)$
$=\left(\alpha^{3}\right)\left(\beta^{3}\right)=(1)(1)=1$
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