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?

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: $\sqrt[4]{3}+\sqrt[4]\frac{1}{27}=\sqrt[4]{3} \cdot \frac{4}{3}$.
