Let $$f$$ be a real valued continuous function defined on the positive real axis such that $$g(x)=\int_{0}^{x}t f(t)dt$$. If $$g(x^{3})=x^{6}+x^{7}$$, then Value of $$\sum_{r=1}^{15}f(r^{3})$$ is:

Given $$g(x) = \int_0^x tf(t)dt$$ and $$g(x^3) = x^6 + x^7$$.

Differentiating both sides with respect to $$x$$:

$$g'(x^3) \cdot 3x^2 = 6x^5 + 7x^6$$

Since $$g'(x) = xf(x)$$: $$x^3 f(x^3) \cdot 3x^2 = 6x^5 + 7x^6$$.

$$3x^5 f(x^3) = x^5(6 + 7x)$$, so $$f(x^3) = 2 + \frac{7x}{3}$$.

$$\sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15}\left(2 + \frac{7r}{3}\right) = 30 + \frac{7}{3} \cdot \frac{15 \times 16}{2} = 30 + 280 = 310$$.

The correct answer is Option 4: 310.