Let a and b are two positive real numbers such that a² + b = 2, then the maximum value of term independent of x in the expansion of ${\left({\mathrm{ax}}^{\frac{1}{6}}+{\mathrm{bx}}^{\frac{-1}{3}}\right)}^9$is ______

${\mathrm{T}}_{\mathrm{r}+1}{=}^9{\mathrm{C}}_{\mathrm{r}}{\mathrm{a}}^{\mathrm{a}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}}{\mathrm{x}}^{\frac{3-\mathrm{r}}{2}}$

So, for term independent of x, r = 3.

$\therefore$Term independent of $\mathrm{x}{=}^9{\mathrm{C}}_3{\mathrm{a}}^6{\mathrm{b}}^3$

Now using $\text{ A.M }\ge \text{ G.M. }$

$\begin{array}{l}\frac{3\left(\frac{{\mathrm{a}}^2}{3}\right)+3\left(\frac{\mathrm{b}}{3}\right)}{6}\ge {\left[\frac{{\mathrm{a}}^6{\mathrm{b}}^3}{3^6}\right]}^{\frac{1}{6}}\\ \Rightarrow {\mathrm{a}}^6{\mathrm{b}}^3\le 1\end{array}$

So, maximum value of term independent of x is $9{\mathrm{C}}_3=84$