Piece of cake part 2

A little manipulation on AM-GM, first we call the expression $P$ , set $t=2x+1$ and $P$ becomes $P=t^5\bigg(\frac{5-3t}{2}\bigg)^3\Rightarrow 2^3P=t^5(5-3t)^3$ So now we need to clear out $t$ on the AM side when applying AM-GM. We need a number $a$ such that $5at=9t$ , therefore $a=\frac{9}{5}$ . Then we'll have $\frac{9^5.2^3}{5^5}P=\frac{9^5}{5^5}t^5(5-3t)^3$ Now we can apply AM-GM $2^3\sqrt[8]{\frac{9^5.2^3}{5^5}P}\leq 5.\frac{9}{5}t+3(5-3t)=3.5$ $\Leftrightarrow\frac{3^{10}.2^3}{5^5}P\leq\frac{3^8.5^8}{2^{24}}$ $\Leftrightarrow P\leq\frac{5^{13}}{3^2.2^{27}}=5^{13}.2^{-27}.3^{-2}$ The equality holds when $\frac{9}{5}t=5-3t\Leftrightarrow x=\frac{1}{48}$ . The sum is $5+13+2+3-27-2=-6$