Some equations from Russian maths book. Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2)} + \frac{8}{(x-1)(x+4)} = 1
$$
$$
\sqrt[7]{ (ax-b)^{3}} - \sqrt[7]{ (b-ax)^{3} } =  \frac{65}{8};  a  \neq 0
$$

First question:
Let $y=(x-2)(x-3)$. So it's equivalent to
$$\frac{y(y-2)}{(y+10x)(y+10x-2)}=1$$
$$y^2-2y=y^2+20xy+100x^2-2y-20x=0$$
$$20x=100x^2+20xy$$
For $x\neq 0$,
$$1=5x+(x-2)(x-3)$$
$$x^2=-5$$
$$x=\pm\sqrt{5} i$$
Otherwise $x=0$.
