Question

Let $\omega$ being cube root of unity. Also let, $x=a+b\omega +c{\omega }^2,y=a\omega +b{\omega }^2+c,z=a{\omega }^2+b+c\omega$. Find the value of $\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}$.

Answer 1

Given,

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

Now $x+y+z=a(1+\omega +{\omega }^2)+b(\omega +{\omega }^2+1)+c({\omega }^2+1+\omega )=a.0+b.0+c.0=0$ [ Since $1+\omega ={\omega }^2=0$]

So we have $x+y+z=0$

or,$x+y=-z$......(1).

Now cubing both sides we get,

$x^3+y^3+3xy(x+y)=-z^3$

or, $x^3+y^3+z^3=3xyz$......(2).

Now,

$\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$

$=\frac{x^3+y^3+z^3}{xyz}$

$=\frac{3xyz}{xyz}$ [ Using (2)]

$=3$.