# Find the number of real solutions of the equation $x^3+1=2(2x-1)^{\frac{1}{3}}$

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.

In your original problem, you are allowed. You have $$f(x)=f^{-1}(x)$$, which implies $$f(f(x))=x$$ (assuming both sides lie in domain). Then for increasing function $$f(x)$$, the $$f(f(x))=x$$ implies $$f(x)=x$$. In your example $$(x^3+1)/2$$ is increasing so this approach applies. For simple proof of why this works, I refer you to Mario Carneiro's answer in Solving the equation $f^{-1}(x)=f(x)$.
On the other hand, you cannot apply it on $$f(x)=1/x$$ as you have already found, and this is because it is not an increasing function.
Raise to the power three and factorizing we obtain $$(x-1) \left(x^2+x-1\right) \left(x^6+2 x^4+2 x^3+4 x^2+2 x+9\right)=0$$