Second Solution: IMO Solving Equation Problem: Solve the equation $x+a^3=\sqrt[3]{a-x}$ where a is real.

Solve the equation $x+a^3=\sqrt[3]{a-x}$ where a is real parameter.

My solution:

By observation, note that $x=a-a^3$ is a real solution for the equation $x+a^3=\sqrt[3]{a-x}$.

Now, if we rewrite the given equation by raising it the to the third power and rearrange the terms in descending powers of $x$ and factor it since $x=a-a^3$ is a real solution, we have

$x^3+3a^3x^2+(3a^6+1)x+a(a^8-1)=(x-a+a^3)(x^2+kx+m)$ where $k,\,m$ are constants.

Equating the constant terms from both sides gives $m=\dfrac{a(a^8-1)}{a(a^2-1)}=a^4+a^4+a^2+1$

Equating the coefficients of powers of $x^2$ gives $k=2a^3+a$.

Hence,
$\small x^3+3a^3x^2+(3a^6+1)x+a(a^8-1)=(x-a+a^3)(x^2+(2a^3+a)x+a^4+a^4+a^2+1)$

And the quadratic formula tells us the other two complex roots for the original equation are

$x=\dfrac{-(2a^3+a)\pm\sqrt{-3a^2-4}}{2}$.