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.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2)} + \frac{8}{(x-1)(x+4)} = 1
$$
$$
\sqrt[7]{ (ax-b)^{3}} - \sqrt[7]{ (b-ax)^{3} } =  \frac{65}{8};  a  \neq 0
$$
 A: 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 $x$.
A: 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 $x=0$.
A: 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.
A: For the first one, consider the equation as $$\frac{f(x)}{g(x)}=1$$
However $$f(-x)\equiv g(x)$$
Therefore $f$ and $g$ are reflections of each other in the $y$ axis.
Neither has symmetry in the $y$ axis, and neither have roots of the form $x=\pm a$, so the only intersection is on the $y$ axis, so $x=0$ is the only solution.
A: factorizing the left minus the right-hand side we get $$-\frac{20 x \left(x^2+5\right)}{(x+1) (x+2) (x+3) (x+4)}=0$$
can you proceed from here?
making the same with the second equation we get $$-\frac{x (x+3) \left(x^2+3 x-16\right)}{(x-1) (x+1) (x+2) (x+4)}=0$$
