Question

If $p,q,r$ are positive integers and $\omega$ be an imaginary cube root of unity and $f\left(x\right)=x^{3p}+x^{3q+1}+x^{3r+2}$, then $f\left(\omega \right)$ is equal to

• A. $1$
• B. $0$
• C. $-1$
• D. $Noneofthese$

Answer 1

The correct option is A $1$

Given $f\left(x\right)=x^{3p}+x^{3q+1}+x^{3r+2}$

$f\left(x\right)=x^{3p}+x^{3q+1}+x^{3r+2}$

$f\left(\omega \right)={\left(\omega \right)}^{3p}+{\left(\omega \right)}^{3q+1}+{\left(\omega \right)}^{3r+2}$

$f\left(\omega \right)={\left({\omega }^3\right)}^p+{\left({\omega }^3\right)}^q\omega +{\left({\omega }^3\right)}^r{\omega }^2$

$f\left(\omega \right)=1+\omega +{\omega }^2$

$f\left(\omega \right)=0$