Question:
Find the number of real solutions of the following equation: $$x^3+1=2(2x-1)^{\frac{1}{3}}$$
My Approach:
Since the question was under the topic "Inverse Functions", I just tried to relate this problem to the mentioned topic. Luckily, I found $\frac{x^3+1}{2}$ and $(2x-1)^{\frac{1}{3}}$ are inverses of each other, by using the general method of expressing $x$ in terms of $y$ for either L.H.S. or R.H.S.
We know a function and its inverse are mirror images about the line $y=x$. So I thought, the function and it's inverse if at all they meet or intersect, they must do so on the line $y=x$.
So, I solved $\frac{x^3+1}{2}=x$ and obtained the values of $x$ as follows: $$x=1, \frac{-1+\sqrt5}{2},\frac{-1-\sqrt5}{2}$$
So there are three values of $x$ that satisfy the obtained equation and so the answer for the above question, i.e., number of real solutions is three. And the answer is also correct according to the book.
Then, I thought of functions like $y=\frac{1}{x}$ which intersect with their inverses at points other than on the line $y=x$. So the above method I used is not rigorous even though it gave the correct answer here.
Kindly give me an alternate solution for this problem and tell me whether I am allowed to use the method described above.