Let $$f(x)$$ and $$g(x)$$ be two functions satisfying $$f(x^2) + g(4-x) = 4x^3$$ and $$g(4-x) + g(x) = 0$$, then the value of $$\int_{-4}^{4} f(x^2) dx$$ is ________.

We are given $$f(x^2) + g(4-x) = 4x^3$$ $$-(1)$$ and $$g(4-x) + g(x) = 0$$ $$-(2)$$.

From equation $$(2)$$, $$g(4-x) = -g(x)$$. Substituting into equation $$(1)$$: $$f(x^2) = 4x^3 + g(x)$$ $$-(3)$$.

Since $$f(x^2)$$ depends only on $$x^2$$, replacing $$x$$ by $$-x$$ in equation $$(3)$$: $$f(x^2) = -4x^3 + g(-x)$$ $$-(4)$$.

From equations $$(3)$$ and $$(4)$$: $$4x^3 + g(x) = -4x^3 + g(-x)$$, so $$g(-x) - g(x) = 8x^3$$ $$-(5)$$.

Adding equations $$(3)$$ and $$(4)$$: $$2f(x^2) = g(x) + g(-x)$$, so $$f(x^2) = \frac{g(x) + g(-x)}{2}$$ $$-(6)$$.

Now we compute $$\int_{-4}^{4} f(x^2)\, dx$$. Since $$f(x^2)$$ is an even function of $$x$$ (because $$f((-x)^2) = f(x^2)$$), we have $$\int_{-4}^{4} f(x^2)\, dx = 2\int_0^4 f(x^2)\, dx$$.

Using equation $$(3)$$: $$2\int_0^4 f(x^2)\, dx = 2\int_0^4 [4x^3 + g(x)]\, dx = 2\int_0^4 4x^3\, dx + 2\int_0^4 g(x)\, dx$$.

The first part: $$2\int_0^4 4x^3\, dx = 2 \cdot [x^4]_0^4 = 2 \cdot 256 = 512$$.

For the second part, we use the substitution $$x = 4 - t$$ in $$\int_0^4 g(x)\, dx$$. This gives $$\int_4^0 g(4-t)(-dt) = \int_0^4 g(4-t)\, dt$$. From equation $$(2)$$ with $$x = t$$: $$g(4-t) = -g(t)$$. So $$\int_0^4 g(x)\, dx = \int_0^4 (-g(t))\, dt = -\int_0^4 g(t)\, dt$$.

This means $$\int_0^4 g(x)\, dx = -\int_0^4 g(x)\, dx$$, which gives $$2\int_0^4 g(x)\, dx = 0$$, so $$\int_0^4 g(x)\, dx = 0$$.

Therefore $$\int_{-4}^{4} f(x^2)\, dx = 512 + 2(0) = 512$$.