Maxima-Minima?

Let $(x-2)=a$ and $(9-x)=b$ ,

Thus, $a+b=7$

or $\large \frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{4}+\frac{b}{4}+\frac{b}{4}+\frac{b}{4}=7$

To calculate the maximum value of $a^{3}b^{4}$ ,

Applying $\text{AM} \geq \text{GM}$ ,

$\large \large \frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{4}+\frac{b}{4}+\frac{b}{4}+\frac{b}{4}}{7} \geq (\frac{a^{3}b^{4}}{3^{3}4^{4}})^{\frac{1}{7}}$

$a^{3}b^{4} \leq 3^{3}4^{4} = 6912$

$a^{3}b^{4} \leq 6912$

$(x-2)^{3}(9-x)^{4} \leq \boxed{6912}$