Let $x,y,z$ be positive real numbers such that $x+y+z=12$ and $x^3{y}^4{z}^5=\left(0.1\right){\left(600\right)}^3$. Then $x^3+y^3+z^3$ is equal to

$x+y+z=12$

Now, $A.M\ge G.M.$

$\Rightarrow \frac{3\left(\frac{x}{3}\right)+4\left(\frac{y}{4}\right)+5\left(\frac{z}{5}\right)}{12}\ge {\left[{\left(\frac{x}{3}\right)}^3{\left(\frac{y}{4}\right)}^4{\left(\frac{z}{5}\right)}^5\right]}^{1/12}$

$\Rightarrow \frac{x^3{y}^4{z}^5}{3^3{4}^4{5}^5}\le 1\Rightarrow x^3{y}^4{z}^5\le 3^3{.4}^4{.5}^5$

But,given $x^3{y}^4{z}^5=\left(0.1\right){\left(600\right)}^3$

$\therefore$ all the numbers are equal.

$\therefore \frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(say\right)\Rightarrow x=3k,y=4k,z=5k$

But, $x+y+z=12\Rightarrow 3k+4k+5k=12\Rightarrow k=1$

$\therefore x=3;y=4;z=5So,x^3+y^3+z^3=216$