How to understand for which $a, b\in\mathbb{R}$ the equation $a-x+\frac{b}{x^3}=0$ has a unique zero? Let $a, b\in\mathbb{R}$, $x\in\mathbb{R}^*$ and consider te he equation
$$a-x+\frac{b}{x^3}=0.$$
My question is: there is a way to understand for which values of $a, b$ has a unique zero?
I tried by using wolfram, but I am not so practice with that.
Thank you in advance!
${\bf EDIT:}$ Availing of the comment of David P, I have
$$a-x+\frac{b}{x^3}=0 \iff ax^3-x^4 +b=0 \iff x^4-ax^3-b=0.$$
Now, let $f(x) =x^4-ax^3-b$. It is $f^{\prime}(x) = 4x^3 -3ax^2$ and then
$$f^{\prime}(x) =0\iff x^2(4x-3a)=0\iff x=\frac34 a.$$
On the other hand, it is
$$f(\frac34 a) = 0\iff \frac14 \left(\frac34\right)^3 a^4 +b=0\iff b= -\frac14 \left(\frac34\right)^3 a^4.$$
Could someone please help me to check the signs of $f$ aside from the zero?
 A: Let $y=a-x+\frac{b}{x^3}$
First consider case $b > 0$. Then $y(x\to -\infty)=+\infty$, $y(x\to -0)=-\infty$, $y(x\to +0)=+\infty$, $y(x\to +\infty)=-\infty$. So there are at least two zeroes: one in $(-\infty;0)$ and second in $(0;+\infty)$. So this case is not consistent with condition.
Consider case $b=0$. Then the only zero is $x=a$. This case is consistent with condition.
Consider case $b < 0$. Then $y(x\to -\infty)=+\infty$, $y(x\to -0)=+\infty$, $y(x\to +0)=-\infty$, $y(x\to +\infty)=-\infty$. $y$ as function of $x$ has minimum in $(-\infty;0)$ and has maximum in $0;+\infty)$. One can use derivatives to find extremums:
$$y'=-1-\frac{3b}{x^4}=0 \Rightarrow x=\pm\sqrt[4]{-3b}$$
Values of $y$ in extremum points: $y_1=y(-\sqrt[4]{-3b})=a+\sqrt[4]{-3b}-\frac{b}{\sqrt[4]{-27b^3}}=a+\sqrt[4]{-b}\cdot(\sqrt[4]{3}+\sqrt[4]\frac{1}{27})$, $y_2=y(\sqrt[4]{-3b})=a-\sqrt[4]{-b}\cdot(\sqrt[4]{3}+\sqrt[4]\frac{1}{27})$.
For zero to be unique there are two possibilities: $y_1=0$, $y_2<0$ or $y_1>0$, $y_2=0$. As $y_1>y_2$ at $b < 0$ it is sufficient $y_1=0$ for first case and $y_2=0$ for second case. Then there are two possible values of $a$ for every $b < 0$: $a=\pm \sqrt[4]{-b}\cdot(\sqrt[4]{3}+\sqrt[4]\frac{1}{27})$.
Update: $\sqrt[4]{3}+\sqrt[4]\frac{1}{27}=\sqrt[4]{3} \cdot \frac{4}{3}$.
A: To take a graphical interpretation of the problem, the equation $ \  \frac{b}{x^3} \ = \ x-a \ \ $  can be taken as providing the $ \ x-$coordinates of intersection(s) of the curve $ \ y \ = \ \frac{b}{x^3} \ $ with a line of slope $ \ 1 \ \ , \ y \ = \ x - a \ \ , \ $ if they exist.
The family of curves  representing $ \ y \ = \ \frac{b}{x^3} \ \ $ will (generally) have two "branches", appearing similar to the rectangular hyperbolas $ \ y \ = \ \frac{b}{x} \ \ , \ $ but hewing more tightly to the coordinate axes.  The curves have symmetry about the origin (if $ \ (x,y) \ $ lies on a particular curve, so does $ \ (-x \ , \ -y) \ ) \ . \ $  For $ \ b \ > \ 0 \ \ , \ $ the "branches" of the curve are found in the first and third quadrants, which necessarily means that a line $ \ y \ = \ x - a \ $ must have an intersection in each of those quadrants for any (real) value of $ \ a \ \ . \ $  The "trivial" case occurs for $ \ b \ = \ 0 \ \ , \ $ where the curve "degenerates" to the $ \ x-$axis and there is a single root of the equation $ \ ( x \ = \ a ) \ $ for any real $ \ a \ \ . \ $
With $ \ b \ < \ 0 \ \ , \ $ the curve  $ \ y \ = \ \frac{b}{x^3} \ \ $ has its two branches in the second and fourth quadrants, which now means that there is an interval of values of $ \ a \ $ for which $ \ y \ = \ x - a \ $ has no intersections at all.  On the other hand, it is clear that a line of slope $ \ 1 \ $ will intersect the curve at two points for large values of $ \ |a| \ $ (either in the second quadrant for $ \ a \ < \ 0 \ \ , \ $ or in the fourth for  $ \ a \ > \ 0 \ \ . \ ) \ . \ $  We can locate the points of tangency of the lines with $ \ y \ = \ \frac{-|b|}{x^3} \ \ , \  $ which serve to define the "gap" between the two branches, by solving
$$  \frac{d}{dx}   \left[ \ \frac{-|b|}{x^3} \ \right] \ \ = \ \ \frac{3·|b|}{x^4} \ \ = \ \ 1 \ \ \Rightarrow \ \ x \ \ = \ \ \pm \ \sqrt[4]{3·|b|} \ \ , $$
$$ y \ \ = \ \ \frac{-|b|}{\pm \ (3·|b|)^{3/4}} \ \ = \ \ \mp \ \sqrt[4]{\frac{|b|}{27}} \ \ , \ \  $$
indicating the symmetry about the origin of the two tangent points.  These points lie on the separate lines
$$ y \ \pm \ \sqrt[4]{\frac{|b|}{27}} \ \ = \ \ x \ \mp \ \sqrt[4]{3·|b|} \ \ \Rightarrow \ \ y \ \ = \ \ x \ \mp \ \left( \ \sqrt[4]{\frac{81·|b|}{27}} \ + \ \sqrt[4]{\frac{ |b|}{27}} \ \right) $$
$$ = \ \ x \ \mp \ \left( \    \sqrt[4]{ \frac{ |b|}{27}}  \ ·  \ [ \ 3  \ + \ 1 \ ] \  \right) \ \ = \ \ x \ \mp \ 4·\left( \    \sqrt[4]{ \frac{ |b|}{27}}  \  \right) \ \ . $$
We thus find the $ \ y-$intercepts of the two tangent lines for $ \ b \ < \ 0 \ $ to be $ \ \large{ a \ = \ \pm \ 4·\left( \ \sqrt[4]{ \frac{ |b|}{27}}  \ \right) } \ \ , \ $ and conclude that the equation $ \ a - x + \frac{b}{x^3} \ = \ 0 \ $ has a unique zero for $ \ b \ = \ 0 \ $ (trivial) and for
$$  |b|  \ \ = \ \ 27·\left(\frac{a}{4} \right)^4 \ \ , \ \ b \ < \ 0 \ \ , $$
confirming your result.
