I think this is swifty

It is easy to see that $x = 0$ satisfy the equation. To prove that this is unique, the equation...

$\sqrt{x + 3} - \sqrt{3-x} + \sqrt{x + 2} - \sqrt{2-x} + \sqrt{x+1} -\sqrt{1-x} = 0$

implies $\sqrt{x + 3} + \sqrt{x + 2} + \sqrt{x+1} = \sqrt{3-x} +\sqrt{2-x} + \sqrt{1-x}$

Noticed that the LHS and RHS are both one-to-one and LHS is increasing while RHS is decreasing. Thus, the equation has at most 1 solution. Therefore, the sum of the roots is $0$ .

After changing $x$ to $-x$ and some algebraic manipulation, a same equation is formed.
This means that if $x$ is a solution, so does $-x$ .
The sum $=0$

The equation above is equivalent to $\frac{x}{\sqrt{x+3}+\sqrt{3-x}}+\frac{x}{\sqrt{x+2}+\sqrt{2-x}}+\frac{x}{\sqrt{x+1}+\sqrt{1-x}}=0$ $x\bigg(\frac{1}{\sqrt{x+3}+\sqrt{3-x}}+\frac{1}{\sqrt{x+2}+\sqrt{2-x}}+\frac{1}{\sqrt{x+1}+\sqrt{1-x}}\bigg)=0$ Since $\frac{1}{\sqrt{x+3}+\sqrt{3-x}}+\frac{1}{\sqrt{x+2}+\sqrt{2-x}}+\frac{1}{\sqrt{x+1}+\sqrt{1-x}}>0; \forall x\in [-1;1]$ so the only solution is $x=0$