Let $\mathrm{\alpha }>0, \mathrm{\beta }>0$ be such that ${\mathrm{\alpha }}^3+{\mathrm{\beta }}^2=4$. If the maximum value of the term independent of x in the binomial expansion of ${\left({\mathrm{\alpha x}}^{1/9}+{\mathrm{\beta x}}^{-1/6}\right)}^{10}$ is 10k, then k =

In the expansion ${\left({\mathrm{\alpha x}}^{1/9}+{\mathrm{\beta x}}^{-1/6}\right)}^{10},{\mathrm{T}}_{\mathrm{r}+1}{=}^{10}{\mathrm{C}}_{\mathrm{r}}{\left({\mathrm{\alpha x}}^{1/9}\right)}^{10-\mathrm{r}}{\left({\mathrm{\beta x}}^{-1/6}\right)}^{\mathrm{r}}{=}^{10}{\mathrm{C}}_{\mathrm{r}}{\mathrm{\alpha }}^{10-\mathrm{r}}{\mathrm{\beta }}^{\mathrm{r}}{\mathrm{x}}^{\frac{10-\mathrm{r}}{9}-\frac{\mathrm{r}}{6}}$

If $\frac{10-\mathrm{r}}{9}-\frac{\mathrm{r}}{6}=0$ then $\frac{10-\mathrm{r}}{9}=\frac{\mathrm{r}}{6}\Rightarrow 60-6\mathrm{r}=9\mathrm{r}\Rightarrow 15\mathrm{r}=60\Rightarrow \mathrm{r}=4$.

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

A.M of ${\mathrm{\alpha }}^3,{\mathrm{\beta }}^2\ge$G.M. of ${\mathrm{\alpha }}^3,{\mathrm{\beta }}^2\Rightarrow \frac{{\mathrm{\alpha }}^3+{\mathrm{\beta }}^2}{2}\ge \sqrt{{\mathrm{\alpha }}^3{\mathrm{\beta }}^2}\Rightarrow \sqrt{{\mathrm{\alpha }}^3{\mathrm{\beta }}^2}\le \frac{{\mathrm{\alpha }}^3+{\mathrm{\beta }}^2}{2}=\frac{4}{2}=2$

$\Rightarrow {\mathrm{\alpha }}^3{\mathrm{\beta }}^2\le 4\Rightarrow {\mathrm{\alpha }}^6{\mathrm{\beta }}^4\le 16$

$\therefore$Maximum value of ${\mathrm{T}}_5$ is ${ }^{10}{\mathrm{C}}_4\left(16\right)=\frac{10\times 9\times 8\times 7}{1\times 2\times 3\times 4}\left(16\right)=210\times 16$

$\therefore$$10\mathrm{k}=210\times 16\Rightarrow \mathrm{k}=21\times 16=336$