Evaluate $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$ 
If $g\left( x \right) = ax + c$, $f\left( x \right) = {x^b} + 3$, and $(g \circ f)^{-1}(x) = {\left( {\frac{{x - 7}}{2}} \right)^{\frac{1}{3}}}$ what is the value of $a+b+c$?

My approach is as follows
Given $g\left( x \right) = ax + c$ & $f\left( x \right) = {x^b} + 3 \Rightarrow {\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = x$
$\frac{{g\left( x \right) - c}}{a} = x$
${\left( {f\left( x \right) - 3} \right)^{\frac{1}{b}}} = \frac{{g\left( x \right) - c}}{a}$
$g\left( x \right) = g\left( x \right) \Rightarrow {g^{ - 1}}\left( {g\left( x \right)} \right) = x$
I am not able to proceed.

Hint: Recall that $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$. You've already compute each inverse separately, so now we just compose them to get:
$$
(g \circ f)^{-1}(x) = f^{-1}(g^{-1}(x))
= f^{-1}\left(\frac{x - c}{a}\right)
= \left( \left(\frac{x - c}{a}\right) - 3 \right)^{\frac{1}{b}}
= \left( \frac{x - 3a - c}{a}\right)^{\frac{1}{b}}
$$
