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: Consider the derivative of $f(x)=1+a^2+ax-x^3$, that is
$$
f'(x)=a-3x^2
$$
This vanishes at $x=\pm\sqrt{a/3}$ and we can compute the values at the relative minimum $-\sqrt{a/3}$; it's simpler if we set $\sqrt{a/3}=b$, so $a=3b^2$:
$$
f(-b)=1+9b^4-3b^3+b^3=1-2b^3+9b^4
$$
The function decreases from $\infty$ to the relative minimum then, goes up to the relative maximum and then crosses once the $x$-axis decreasing to $-\infty$. So, if we prove that $f(-b)>0$, we're done.
Consider $g(x)=1-2x^3+9x^4$; its derivative is $g'(x)=-6x^2+36x^3=6x^2(6x-1)$, so the function has an absolute minimum at $x=1/6$. Since
$$
g(1/6)=1-\dfrac{1}{108}+\dfrac{1}{144}=\dfrac{431}{432}>0
$$
we're done.
By the way, for $a\le0$ the function is globally decreasing, so the equation has a single real solution also in this case.
A: 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 $a$.
A: 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.
A: Just to throw in another way to do this, since the equation has no $x^2$ term, the sum of its roots must be zero.  If it has multiple real roots, then they are all real and (to add up to zero) at least one root must be negative.
Complete a square for $a$ and rearrange:
$$(a+x/2)^2 = x^3 + \frac{x^2}{4} - 1$$
Basic analysis for RHS shows that it's negative when $x\le0$, so this equation cannot have a negative root.
Therefore the equation can only have one real root.
A: Write $f(x)=x^3-ax-a^2-1.$ Then we have that $f'(x)=3x^2-a.$ Hence the function has its local extrema at $$x=\pm\sqrt {\frac a3}.$$ It follows that there is exactly one real root if the relative extrema have equal sign. Hence we compute $$f\left(\sqrt {\frac a3}\right)f\left(-\sqrt {\frac a3}\right)$$ and confirm that it is positive. This gives $$\left[a^2+1 +a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right]\left[a^2+1-\left(a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right)\right]=(a^2+1)^2-\left[a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right]^2=(a^2+1)^2-\frac a3\left(a-\frac a3\right)^2=a^4-\frac{4}{27}a^3+2a^2+1=\left(a^2+\frac{2a}{27}\right)^2+\left(2-\frac{4}{27^2}\right)a^2+1,$$ which is clearly positive for all $a.$
