Question

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

Answer 1

Given that x³ + 4y³ + 9z³ = 18xyz and x + 2y + 3z = 0
$\therefore$ x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y

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

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

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

Given that x³ + 4y³ + 9z³ = 18xyz

$\therefore \frac{(x+2y{)}^2}{xy}+\frac{(2y+3z{)}^2}{yz}+\frac{(3z+x{)}^2}{zx}=\frac{18xyz}{xyz}$

= 18