Solutions for $a$ by factoring a multivariate polynomial I have an equation:
$$\left(\frac{b}{x^2}+1\right)⋅\left(x−\frac bx\right)+a=0$$
The question is by factorizing what are the solutions for a?
I am not sure how to do this: 
I have reduced the equation to:
$$\frac{b^2}{x^3}+x+a=0\\
  x^4 - b^2 + ax^3 = 0 $$
An example question (which is not related to the above), is very difficult to understand and only gives one real solution, but may be useful in those trying to help:
The question states if:
$$x^3+xb+a=0 \\$$
one of the solutions is:
$$a=2\cdot \left(\frac b3\right)^{3/2}$$
How did they arrive at this?

We have the equation: $$\left(\frac b{x^2}+1\right)⋅\left(x−\frac bx\right)+a=0$$
and hence the domain of $x$ must exclude $x = 0$.
$$\begin{align} \left(\frac b{x^2}+1\right)⋅\left(x−\frac bx\right)+a=0 & \iff \frac bx -\frac{b^2}{x^3} + x - \frac bx + a = 0 \\ \\ &\iff x-\frac{b^2}{x^3} +a =0\\ \\& \iff \frac{x^4 - b^2+ ax^3}{x^3}=0\end{align}$$
$$x^4 - b^2 + ax^3 = 0 \iff x - \frac{b^2}{x^3} + a = 0 \implies a = \frac{b^2}{x^3} - x$$

Please note that you erred when you went from 
$$x^4 + b^2 + ax^3 = 0 $$
to $$x^3 + bx + a = 0$$
These are not the same equations. 

Regarding your second question, to determine whether the given $a$ is a solution, plug  that solution into $x^3 + bx + a = 0$.
