Question

If $x=a+b, y=a \omega+b \omega^{2}, z=a \omega^{2}+b \omega$, then the value of $x^{3}+y^{3}+z^{3}$ is equal to (where $\omega$ is imaginary cube root of unity)

• A. $a^{3}+b^{3}$
• B. $3\left(a^{3}+b^{3}\right)$
• C. $3\left(a^{2}+b^{2}\right)$
• D. None of these

Answer 1

Answer: (C)

We have
$x^{3}+y^{3}+z^{3}=(a+b)^{3}+\left(a \omega+b \omega^{2}\right)^{3}+\left(a \omega^{2}+b \omega\right)^{3} $
$=3 a^{3}+3 b^{3}+3\left(a^{2} b+a b^{2}\right)\left(1+\omega^{2} \omega^{2}+\omega \omega^{4}\right)$
$=3 a^{3}+3 b^{3}+3\left(a^{2} b+a b^{2}\right)\left(1+\omega+\omega^{2}\right)=3\left(a^{3}+b^{3}\right)$
Trick: As in the previous question
$x^{3}+y^{3}+z^{3}=(4)^{3}+(-2)^{3}+(-2)^{3}=48$ and $(b)$
i.e., $3\left(a^{3}+b^{3}\right)=48$