An algebra problem by Raghav Rathi

The equation is equivalent to $\dfrac{x^3+1}2 = \sqrt[3]{2x-1}$ . Let $f(x) = \dfrac{x^3 + 1} 2$ , then the inverse function $f^{-1}(x)$ exists, and is equal to $\sqrt[3]{2x-1}$ . Hence, our equation in question can be written as $f(x) = f^{-1}(x)$ .

Since a function $f$ is equal to its inverse, then the solution (if any) occurs at the intersection point(s) between the function $f$ and the straight line $y=x$ . Thus, we just need to solve for $f(x) = x\qquad \Leftrightarrow \qquad \dfrac{x^3 + 1}2 = x\qquad \Leftrightarrow \qquad x^3 - 2x + 1 = 0.$

By rational root theorem , we can see that $x = 1$ is a solution. Factoing out this linear factor gives $(x-1)(x^2 + x -1 ) = 0$ . What remains to solve is the quadratic equation, applying the quadratic formula gives the other two roots, namely, $x = \dfrac{\sqrt5 \pm 1 } 2$ . However, upon inspection, we see that $x =\dfrac{\sqrt5 +1}2$ does not satisfy the original equation. Hence, the solution set to the equation in question is $\boxed{ \left \{ 1 , \frac{\sqrt5 -1}2 \right \}}$ .

Footnote : Note that $f^{-1}(x)$ exists because $f(x)$ is a one-to-one function.