Raja Oktovin wrote: Solve the equation \[ x + a^3 = \sqrt [3]{a - x}\] where $ a$ is a real parameter. Let $ x=y-a^3$. The equation becomes $ y=\sqrt[3]{a+a^3-y}$ $ \implies$ $ y^3=a+a^3-y$ $ \iff$ $ (y-a)(y^2+ay+a^2+1)=0$ $ \iff$ $ (y-a)((y+\frac a2)^2+\frac{3a^2+4}4)=0$ Hence the unique real solution $ x=a-a^3$

hello, i think the solution must be $ x=a-a^3$ and $ a\geq0$. Sonnhard.

Dr Sonnhard Graubner wrote: hello, i think the solution must be $ x = a - a^3$ and $ a\geq0$. Sonnhard. but why must $ a \ge 0$? if $ a=-1$, then clearly $ x=0$ since the cubic root can operate in negative number?