<s>[INST] Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$ Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$

Let $P(x)=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$
By using $AM-GM$ inequality on the first two terms give me
$8(27)^{\log_{6}x}+27(8)^{\log_{6}x}\geq2\sqrt{8\times 27\times(27\times 8)^{\log_6x}}\geq 12\sqrt{6x^3}$
$P(x)=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3\geq12\sqrt{6x^3}-x^3$
I found the maximum value of $12\sqrt{6x^3}-x^3$ by differentiating and equating it to zero.I found that $12\sqrt{6x^3}-x^3$ is maximum at $x=6$.And its maximum value is $216$.
Can the maximum value of $12\sqrt{6x^3}-x^3$ be the maximum value of $P(x)?$I am confused because P(x)is greater than or equal to $12\sqrt{6x^3}-x^3$.$P(x)$ may be greater than the maximum value of $12\sqrt{6x^3}-x^3$.
 [/INST] Hint:
Let $E=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3=8(3^3)^{\log_{6}x}+27(2^3)^{\log_{6}x}-x^3$
$$=8(3)^{\log_{6}x^3}+27(2)^{\log_{6}x^3}-x^3$$
Now put $\log_{6}x^3=t$
$ \Rightarrow 6^t=x^3$
So we have that   $E=8(3)^{t}+27(2)^{t}-6^t$
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