Question

If $x+2y+3z=0$ and $x^3+4y^3+9z^3=18xyz$;

Evaluate :

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

• A. $21$
• B. $12$
• C. $18$
• D. $7$

Answer 1

The correct option is C $18$

Given, $x+2y+3z=0$ and $x^3+4y^3+9z^3=18xyz$
$=>x+2y=-3z$
also, $=>2y+3z=-x$
and $=>3z+x=-2y$
Now,
$\frac{{(x+2y)}^2}{xy}+\frac{{(2y+3z)}^2}{yz}+\frac{{(3z+x)}^2}{zx}$
$=\frac{{(-3z)}^2}{xy}+\frac{{(-x)}^2}{yz}+\frac{{(-2y)}^2}{zx}$
$=\frac{9z^2}{xy}+\frac{x^2}{zy}+\frac{4y^2}{zx}$
$=\frac{9z^3}{xyz}+\frac{x^3}{xyz}+\frac{4y^3}{xyz}$
$=\frac{9z^3+x^3+4y^3}{xyz}$
$=\frac{18xyz}{xyz}$
$=18$