Question

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

Answer 1

$x=a+b\omega +c{\omega }^2,y=a\omega +b{\omega }^2+c$

$z=a{\omega }^2+b+c\omega$

$x+y+z=a(1+\omega +{\omega }^2)+b(\omega +{\omega }^2+1)+c({\omega }^2+1+\omega )$

$=a\left(0\right)+b\left(0\right)+c\left(0\right)$

$x+y+z=0(\because 1+\omega +{\omega }^2=0)$

If $x+y+z=0$, then we know that $x^3+y^3+z^3=3xyz$

$\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=\frac{x^3+y^3+z^3}{xyz}=\frac{3xyz}{xyz}=3$

$\therefore \frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3$