Show that there is only one real root of the equation $1 + a^2 + ax - x^3$ I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?
Thanks.

A cubic has only one real root iff its discriminant is negative. (See Wikipedia.)
The discriminant of $1 + a^2 + ax - x^3$ is $-27 a^4 + 4 a^3 - 54 a^2 - 27$.
Now, $-27 a^4 + 4 a^3 - 54 a^2 = a^2 (-27 a^2 + 4 a - 54)$ and $-27 a^2 + 4 a - 54=-27 \left(a - \frac{2}{27}\right)^2 - \frac{1454}{27} < 0$ for all $a$.
Therefore, $-27 a^4 + 4 a^3 - 54 a^2 - 27 < 0$ for all $a$.
