For some $$a, b, c \in \mathbb{N}$$, let $$f(x) = ax - 3$$ and $$g(x) = x^b + c$$, $$x \in \mathbb{R}$$. If $$(f \circ g)^{-1}(x) = \left(\frac{x-7}{2}\right)^{1/3}$$, then $$(f \circ g)(ac) + (g \circ f)(b)$$ is equal to _____.

Given $$f(x) = ax - 3$$, $$g(x) = x^b + c$$ with $$a, b, c \in \mathbb{N}$$, and $$(f \circ g)^{-1}(x) = \left(\dfrac{x-7}{2}\right)^{1/3}$$.

Finding $$a, b, c$$:

$$f(g(x)) = a(x^b + c) - 3 = ax^b + ac - 3$$

If $$(f \circ g)^{-1}(x) = \left(\frac{x-7}{2}\right)^{1/3}$$, then:

$$f \circ g(x) = 2x^3 + 7$$

Comparing coefficients: $$a = 2$$, $$b = 3$$, $$ac - 3 = 7 \implies c = 5$$.

Computing $$(f \circ g)(ac)$$:

$$ac = 2 \times 5 = 10$$

$$(f \circ g)(10) = 2(10)^3 + 7 = 2000 + 7 = 2007$$

Computing $$(g \circ f)(b)$$:

$$f(3) = 2(3) - 3 = 3$$

$$g(3) = 3^3 + 5 = 32$$

Final answer:

$$(f \circ g)(ac) + (g \circ f)(b) = 2007 + 32 = 2039$$

The answer is $$2039$$.