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

The first equation implies that the product of the distances from $x$ to $-1, -2, -3, -4$ is the same as the product of the distances from $x$ to $1, 2, 3, 4$. (This condition is equivalent to the quotient appearing in the equation being $\pm 1$.) If $x > 0$, the latter distances are smaller than the corresponding former ones, so the latter product is smaller. If $x < 0$, then the former is. So the only possibility is $x = 0$. 
In the second equation, make the substitution $u = x^2 + 3x$.
For the third one, you were almost there with what you wrote in the comments. But remember that $\sqrt[7]{-A} = - \sqrt[7]{A}$. You get $(ax - b)^{3/7} = 65/16$. Raise both sides of the equation to the power of $7/3$.
Edit The third equation has a typo in it. It was supposed to read
$$\sqrt[7]{ (ax-b)^{3}} - \sqrt[7]{ (b-ax)^{-3} } =  \frac{65}{8}.$$
In this case, Write $u = (ax-b)^{3/7}$. Then the equation becomes $u + 1/u = 65/8$. Since after clearing denominators this becomes a quadratic equation, there are at most two possibilities for $u$. Since $u = 8$ and $u = 1/8$ work, these must be the ones. We get $ax - b \in \{128, 1/128\}$, after which it's easy to solve for $x$.
