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