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.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2)} + \frac{8}{(x-1)(x+4)} = 1
$$
$$
\sqrt[7]{ (ax-b)^{3}} - \sqrt[7]{ (b-ax)^{3} } =  \frac{65}{8};  a  \neq 0
$$

1) $\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1\implies\frac{(x^2-5x+4)(x^2-5x+6)}{(x^2+5x+4)(x^2+5x+6)}=1$
$\implies(x^2-5x+4)(x^2-5x+6)=(x^2+5x+4)(x^2+5x+6)$
We know that the even-indexed terms will cancel out because the LHS is just the RHS with $x\to-x$
$\implies-10x^3-50x=10x^3+50x\implies10x^3+50x=0\implies10x(x^2+5)=0\implies x=0$ 
2) Note that the denominators of these can be written as $u+2,u-4$ with $u=x^2+3x$. We see:
$\frac{6}{u+2}+\frac{8}{u-4}=1\implies\frac{6}{u+2}-3+\frac{8}{u-4}+2=1-3+2\implies\frac{-3u}{u+2}+\frac{2u}{u-4}=0$
Remove a factor of $u$ and accept $u=0$ as a solution. Then clear fractions:
$\frac{-3}{u+2}+\frac{2}{u-4}=0\implies-3(u-4)+2(u+2)=0\implies16-u=0\implies u=16$
So, $x^2+3x=0, 16$ - we extract $x=0,-3$ for $u=0$ and solve the $u=16$ case with the quadratic formula / your method of choice.
3) Let $(ax-b)^{3/7}=u$, then $u-(-u)=65/8\implies u=65/16$. Nowhere further to go as $a,b$ aren't given - can say $x=(65/16+b)/a$ if you'd like, I suppose.
Perhaps worth noting that $\frac{65}{8}=8+\frac{1}{8}$, as an aside.
