Question

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

Answer 1

Given $x+2y+3z=0$ __________ (1)

and $x^3+4y^3+a{z}^3=18xyz$ _______ (2)

Now

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

From equation (1) x+2y=-3z, x+3z=2y ; & 2y+3z=-x

then

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

$\Rightarrow \frac{a{z}^2}{xy}+\frac{x^2}{yz}+\frac{4y^2}{zx}=\frac{a{z}^3+x^3+4y^3}{xyz}$

Now from equation (2)

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

$\therefore$ value of equation (3) is equal to 18.