If the maximum value of the term independent of t in the expansion of ${\left({\mathrm{t}}^2{\mathrm{x}}^{\frac{1}{5}}+\frac{(1-\mathrm{x}{)}^{\frac{1}{10}}}{\mathrm{t}}\right)}^{15}, \mathrm{x}\ge 0$, is K, then 8K is equal to _______.

${\left({\mathrm{t}}^2{\mathrm{x}}^{\frac{1}{5}}+\frac{(1-\mathrm{x}{)}^{\frac{1}{10}}}{\mathrm{t}}\right)}^{15}$

${\mathrm{T}}_{\mathrm{r}+1}{=}^{15}{\mathrm{C}}_{\mathrm{r}}{\left({\mathrm{t}}^2{\mathrm{x}}^{\frac{1}{5}}\right)}^{15-\mathrm{r}}\cdot \frac{(1-\mathrm{x}{)}^{\frac{\mathrm{r}}{10}}}{{\mathrm{t}}^{\mathrm{r}}}$

For independent of t,

30 – 2r – r = 0

$\Rightarrow \mathrm{r}=10$

So, Maximum value of $T_{11}={ }^{15}{\mathrm{C}}_{10}x(1-\mathrm{x}){=}^{15}{\mathrm{C}}_{10}\left(\mathrm{x}-{\mathrm{x}}^2\right)$ will be at $\mathrm{x}=\frac{1}{2}$

$$\begin{gathered} at x=\frac{1}{2} \\ k=T_{11}=\frac{15\times 14\times 13\times 12\times 11}{1\times 2\times 3\times 4\times 5}\left(\frac{1}{4}\right)\Rightarrow 8k=6006 \end{gathered}$$