<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] Angelo Mark has provided a good hint, but it seems that the OP needs more. 
We can write
$$\begin{align}8\cdot 3^t+27\cdot 2^t-6^t&=8\cdot 3^t+27\cdot 2^t-2^t\cdot 3^t-8\cdot 27+8\cdot 27\\&=8(3^t-27)-2^t(3^t-27)+8\cdot 27\\&=(3^t-27)(8-2^t)+8\cdot 27\end{align}$$
Note here that $(3^t-27)(8-2^t)$ is non-positive.
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