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

We need to find the maximum value of the term independent of $$t$$ in the expansion of $$\left(t^2 x^{1/5} + \dfrac{(1-x)^{1/10}}{t}\right)^{15}$$, $$x \ge 0$$.

Expanding by the binomial theorem, the general term can be written as:

$$T_{r+1} = \binom{15}{r} \left(t^2 x^{1/5}\right)^{15-r} \left(\dfrac{(1-x)^{1/10}}{t}\right)^r$$

$$= \binom{15}{r} t^{2(15-r)} x^{(15-r)/5} \cdot t^{-r} (1-x)^{r/10}$$

$$= \binom{15}{r} t^{30-2r-r} x^{(15-r)/5} (1-x)^{r/10}$$

$$= \binom{15}{r} t^{30-3r} x^{(15-r)/5} (1-x)^{r/10}$$

For this term to be independent of $$t$$, the exponent of $$t$$ must be zero, which implies

$$30 - 3r = 0 \implies r = 10$$

Substituting $$r = 10$$ into the general term yields the term free of $$t$$ as:

$$T_{11} = \binom{15}{10} x^{(15-10)/5} (1-x)^{10/10} = \binom{15}{5} x(1-x) = 3003 \cdot x(1-x)$$

Thus the problem reduces to maximizing the function $$x(1-x)$$ for $$x \ge 0$$. Applying the AM-GM inequality gives

$$x(1-x) \le \left(\dfrac{x + (1-x)}{2}\right)^2 = \dfrac{1}{4}$$, with equality when $$x = \dfrac{1}{2}$$.

Substituting this maximum into the expression for $$T_{11}$$ gives

$$K = 3003 \cdot \dfrac{1}{4} = \dfrac{3003}{4}$$

$$8K = 8 \cdot \dfrac{3003}{4} = 2 \cdot 3003 = 6006$$

Therefore, the maximum value of the term independent of $$t$$ in the expansion is $$6006$$.