Question

If x + 2y + 3z = 0 and x³ + 4y³ + 9z³ = 18xyz ; evaluate : 
${\displaystyle \frac{{(x+2y)}^2}{xy}+\frac{{(2y+3z)}^2}{yz}+\frac{{(3z+x)}^2}{zx}}$

Answer 1

Given that x³ + 4y³ + 9z³ = 18xyz and x + 2y + 3z = 0
x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y
Now
${\displaystyle \frac{{(x+2y)}^2}{xy}+\frac{{(2y+3z)}^2}{yz}+\frac{{(3z+x)}^2}{zx}}$

= ${\displaystyle \frac{{(-3z)}^2}{xy}+\frac{{(-x)}^2}{yz}+\frac{{(-2y)}^2}{zx}}$

= ${\displaystyle \frac{9z^2}{xy}+\frac{x^2}{yz}+\frac{4y^2}{zx}}$

= ${\displaystyle \frac{x^3+4y^3+9z^3}{xyz}}$

Given that x³ + 4y³ + 9z³ = 18xyz

∴ ${\displaystyle \frac{{(x+2y)}^2}{xy}+\frac{{(2y+3z)}^2}{yz}+\frac{{(3z+x)}^2}{zx}}$

= ${\displaystyle \frac{18xyz}{xyz}}$

= 18