If the maximum value of the term independent of t in the expansion...

Given,

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

Now ${r}^{th}$ term is given by,

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

For independent of $t$,

$2\left(15 - r\right) - r = 0$

$30 - 2r - r = 0$

$⇒ r = 10$

So, Maximum value of $C10 15 x\left(1 - x\right)$ will be at $x = \dfrac{1}{2}\left(\text{as}\dfrac{dy}{dx}\left(x - {x}^{2}\right) = 1 -2 x \text{\text{ and } } 1 - 2x = 0 ⇒ x = \dfrac{1}{2}\right)$

$= C10 15 \dfrac{1}{2} \times \left(\dfrac{1}{2}\right) = \dfrac{3003}{4}$

So, $K = \dfrac{3003}{4}$ and 8K = 6006