Let α > 0 , β > 0 be such that α 3 + β 2 = 4 . If the maximum value of the term…

 $T_{r+1}={}^{10}C_r{\left(\alpha x^{\frac{1}{9}}\right)}^{10-r}{\left(\beta x^{-\frac{1}{6}}\right)}^r$

$T_{r+1}={}^{10}C_r{\alpha }^{10-r}{\beta }^r{\left(x\right)}^{\frac{10-r}{9}-\frac{r}{6}}$

Term independent of $\mathrm{x} \Rightarrow$Power of $x=0$

$\frac{10-r}{9}-\frac{r}{6}=0\Rightarrow r=4$

${\mathrm{T}}_5={}^{10}\mathrm{C}_4{\alpha }^6{\beta }^4$

Now Let ${\alpha }^3,{\beta }^2$ are two numbers. Clearly both are positive.

$A.M.\ge G.M.$

$\Rightarrow \frac{{\alpha }^3+{\beta }^2}{2}\ge {\left({\alpha }^3{\beta }^2\right)}^{1/2}$

$\Rightarrow {\alpha }^3{\beta }^2\le 4$

$\Rightarrow {\alpha }^6{\beta }^4\le 16$

$\Rightarrow \frac{T_5}{{}^{10}{\mathrm{C}}_4}\le 16$

$\Rightarrow T_5\le 16{.}^{10}{\mathrm{C}}_4$

$\Rightarrow T_{5\max }=16{\times }^{10}{\mathrm{C}}_4=10k$

$\Rightarrow 10k=3360$

$\Rightarrow k=336$