Let $p,q,r\in R^{+}$  and $27pqr\ge {\left(p+q+r\right)}^3$  and $3p+4q+5r=12$  then maximum value of $p^3+q^4+r^5=$  ______

$A.M\ge G.M$

$\frac{p+q+r}{3}\ge \sqrt[3]{pqr}$

$p+q+r\ge 3\sqrt[3]{pqr}$

Cubes on both sides

${\left(p+q+r\right)}^3\ge 27pqr\to \left(1\right)$

but given that $27pqr\ge {\left(p+q+r\right)}^3\to \left(2\right)$

${\left(p+q+r\right)}^3=27pqr$

$p+q+r=3\sqrt[3]{pq}$

$\frac{p+q+r}{3}=\sqrt[3]{pqr}$

$A.M=G.M$

$\therefore p=q=r=k$

Now, $3p+4q+5r=12$

$3k+4k+5k=12$

$12k=12$

$k=1$

$p^3+q^4+r^5=1^3+1^4+1^5$

$=3$