Use functions $f(x) =\frac{x}{8} + 1$ and $g(x) = x^3$ to find $(f o g)^-1(5)$ Right answer to the question is
$$
3\sqrt[3]{2}
$$
but i got 
$$
2\sqrt[3]{4}
$$
got it by finding inverse of
$$
\frac{x^3}{8} + 1
$$
and then sustituting 5 into 
$$
2\sqrt[3]{x-1}
$$
who is right?

First of all recall that 
$$\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)={{\left( f\circ g \right)}^{-1}}\left( x \right)$$
it follows that you need to find the inverse functions for both 
we have $f(x)=\frac{x}{8}+1$ then $8f(x)=x+8$ i.e $f^{-1}(x)=8(x-1)$
similarly for $g(x)=x^3$ then $g^{-1}(x)=\sqrt[3]{x}$
hence by above property we get 
${{\left( f\circ g \right)}^{-1}}\left( x \right)=(g^{-1}(f^{-1}(x))=\sqrt[3]{8(x-1)}=2\sqrt[3]{x-1}$
or: you can find $$\left( {{f}}\circ {{g}} \right)\left( x \right)=f(g(x))=\frac{g(x)}{8}+1=\frac{x^3}{8}+1$$
then its easy to find its inverse function  to get 
$${{\left( f\circ g \right)}^{-1}}\left( x \right)=\sqrt[3]{8(x-1)}=2\sqrt[3]{x-1}$$
Seems the error is not $x=5$ to get that result 
