Question

If $x=a+b\omega +c{\omega }^2,y=a\omega +b{\omega }^2+c\&z=a{\omega }^2+b+c\omega$ then the value of $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$ equals

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

From the given, we have
$x+y+z-(a+b\omega +c{\omega }^2)+(a\omega +b{\omega }^2+c)$
$+(a{\omega }^2+b+c\omega )$
$\Rightarrow x+y+z=a(1+\omega +{\omega }^2)+b(1+\omega +c{\omega }^2)$
$+c(1+\omega +{\omega }^2)=0$
$\Rightarrow x^3+y^3+z^3=3xyz$
$\Rightarrow \frac{x^3+y^3+z^3}{xyz}=3$ or $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}=3$