Question

If $\omega \ne 1$ is a cube root of unity and $a+b=21,a^3+b^3=8001$, then the value of $\frac{(a{\omega }^2+b\omega )(a\omega +b{\omega }^2)}{127}$ must be equal to

• A. $3$
• B. $4$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

$3$

$(a{\omega }^2+b\omega )(a\omega +b{\omega }^2)=a^2-ab+b^2$
$\frac{a^3+b^3}{a+b}=\frac{8001}{21}$
$\Rightarrow a^2-ab+b^2=381$
$\therefore \frac{(a{\omega }^2+b\omega )(a\omega +b{\omega }^2)}{127}=\frac{a^2-ab+b^2}{127}=3$