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