<s>[INST] 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.
 [/INST] Notice that $$(g\circ f)(x) = 2x^3+7$$
But by definition of $g$ and $f$ we have also $$(g\circ f)(x) =ax^b+3a+c$$
Now if $x=0$ then we get $7 =3a+c$ and then $ax^b = 2x^3$. So if $x=1$ we get $a=2$ and $c=1$ and $x^b=x^3$. So if $x=2$ we get $2^b = 8$ and thus $b=3$.
Notice: Since we don't know apriori that $b$ is positive integer you can not do just the comparison of the coefficients.
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