<s>[INST] 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.
 [/INST] Oh... for shucks and giggles.
All cubic have at least one real solution.  So let $x=r$ be one real root.
Then $f(x)=(1+a^2) + ax -x^3 = (x-r)(Ax^2 + Bx + C) = Ax^3 + (B-Ar)x^2 + (C-Br)x -Cr$ for some
$A=-1; B+r=0; C-Br= a; -Cr=1+a^2$
So $B = -r; C = -\frac {1+a^2}r$  (assuming $r \ne 0$ but as $f(0) = 1+a^2 \ne 0$ we know $r \ne 0)$.
And $-\frac {1+a^2}r + r^2 = a$.
So $f(x) = (x-r)(-x^2 -rx -\frac {1+a^2}r)$
Any other root of $f(x)$ must be $x = \frac {-r \pm \sqrt{r^2 -4\frac {1+a^2}r}}2=\frac {-r\pm \sqrt{r^2 - 4(r^2 -a)}}2$
But $r^2 \ge 0$ and $a > 0$ so  $r^2 - 4(r^2 -a)=-3r^2 -a=-(r^2 + a)  < 0$. so those solutions can never exist.
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