<s>[INST] Find the inverse of $f(x)=3x^3+1$ I get $\sqrt[3]{\frac{x-1}{3}}$ whereas textbook shows answer as $\sqrt{\frac{x-1}{3}}3$ I am to find the inverse of $f(x)=3x^3+1$. My working:
$$f(x)=3x^3+1$$
$$x=3y^3+1$$
$$x-1=3y^3$$
$$\frac{x-1}{3}=y^3$$
$$f^{-1}(x)=\sqrt[3]\frac{x-1}{3}$$
The solutions section of my textbook says that the answer is  $\sqrt{\frac{x-1}{3}}3$.
How can I arrive at the correct answer? Which step in my working did I go wrong?
 [/INST] This will be a typo, a misplaced $3$
Your answer $f^{-1}(x)=\sqrt[3]{\dfrac{x-1}{3}}$ is correct, as is $\left(\frac{x-1}{3}\right)^{\frac13}$, though you may need to specify using the real cube root rather the principal cube root when $x<1$
If $\sqrt{\frac{x-1}{3}}3$ had really been intended then I suspect it would have been written more tidily as $\sqrt{3({x-1})}$
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