I am trying to show that given non-zero vectors $\mathbf{a}$ and $\mathbf{b}$, the system of equations given by $$\mathbf{a}=\mathbf{x}_2 - \mathbf{x}_1$$ $$\mathbf{b}=\frac{\mathbf{x}_1}{||\mathbf{x}_1||^3}-\frac{\mathbf{x}_2}{||\mathbf{x}_2||^3}$$ can be solved, i.e., a unique solution to the vectors $\mathbf{x}_1$ and $\mathbf{x}_2$ can be found.
Assuming that $\mathbf{x}_1 \neq \mathbf{x}_2$ and $||\mathbf{x}_1|| \neq ||\mathbf{x}_2||$. Also, $||\mathbf{x}_1||$ and $||\mathbf{x}_2||$ are non-zero.
I have a rough idea of how to prove uniqueness for simple nonlinear functions, but this one is a bit tricky. I am not necessarily after the solution to $\mathbf{x}_1$ and $\mathbf{x}_2$, but rather I want to prove that a unique solution does exist (and therefore can be found) given this system of equations.
Edit: I accidentally left out some important information so answers prior to this edit are referring to the old question which states
$$\mathbf{a}=\mathbf{x}_2 - \mathbf{x}_1$$ $$\mathbf{b}=\frac{\mathbf{x}_1}{||\mathbf{x}_1||}-\frac{\mathbf{x}_2}{||\mathbf{x}_2||}$$
Sorry!
The system does not have a unique solution.
The most trivial way would be to say that if $(x_1,x_2)$ solves the equation, then clearly, $(x_2, x_1)$ also solves the equation, but that feels a bit cheap.
But even if you relax your condition and say that switching $x_1,x_2$ does not count as a different solution, the system still does not have a unique solution.
For example, take any $n\in\mathbb N$, and take take $b=0\in\mathbb R^n$, the zero vector, and let $a\mathbb R^n$ be any nonzero vector
Then, let $\lambda \in\mathbb R\setminus\{0\}$ be any nonzero real number.
Then, we can take $x_1=\lambda a$ and $x_2=(\lambda + 1)a$.
We then have
$$x_2-x_1=\lambda a + a - \lambda a = a$$
and $$\frac{x_1}{\|x_1\|} - \frac{x_2}{\|x_2\|} = \frac{\lambda a}{|\lambda|\|a\|} - \frac{(\lambda + 1) a}{|\lambda + 1|\|a\|} = 0 = b$$
so you have an infinite set of values that all solve your two equations.