algebra-precalculus
Problem
Solve the equation $$\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$$
What I've tried
First I tried factoring the denominators but only the second one can be factored as $(x+4)(x-2)$.
Then I tried substituting $y = x^2 – 8$ but that didn't lead me anywhere.
Where I'm stuck
I don't know how to start this problem. Any hints?
P.S. I would really appreciate it if you give me hints or at least hide the solution. Thanks for all your help in advance!
As others have noted, $n=4$ is a natural starting point because of Cauchy's double-then-climb-down inductive strategy. The $n=3$ case is reasonably easy on its own, viz.$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx),$$where the quadratic factor is $\ge0$ by Cauchy-Schwarz, but e.g. $n=7$ isn't as easy as $n=8$. But I disagree with
There are many proofs, none of them are completely straightforward.
In particular, we can use a "normal" proof by induction that doesn't snake through the values of $n$ in a nonstandard way. Perhaps the simplest proof (if the intended audience doesn't know calculus), if we restate the problem as proving $\prod_ia_i=1\implies\sum_ia_i\ge n$ for positive $a_i$, is to arrange $n+1$ terms in the inductive step so $a_1\ge1\ge a_2$, whence$$\begin{align}(a_1-1)(1-a_2)&\ge0\\\implies a_1+a_2&\ge 1+a_1a_2\\\implies\sum_ia_i&\ge 1+\underbrace{a_1a_2+\sum_{i\ge3}a_i}_{\ge n\text{ by inductive hypothesis}}\\&\ge n+1.\end{align}$$Edit: come to think of it, this approach is even simpler, albeit with induction on the number of steps it takes to make all values equal, rather than on $n$.
From
$$\begin{cases} x\sqrt{y} + y\sqrt{x} = 30 & \ \ (a)\\ x\sqrt{x} + y\sqrt{y} = 35& \ \ (b)\end{cases}$$
$3 \times$(a) + (b) gives :
$$(\sqrt{x}+\sqrt{y})^3=5^3\ \ \iff \ \ \sqrt{x}+\sqrt{y}=5 \ \tag{1}$$
Besides, (b) - (a) gives :
$$x(\sqrt{x}-\sqrt{y})-y(\sqrt{x}-\sqrt{y})=5 \ \iff$$
$$(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})=5 \tag{2}$$
Taking (1) into account in (2), on gets :
$$\sqrt{x}-\sqrt{y}=\pm1\tag{3}$$
Gathering (1) and (3), we obtain easily the two solutions :
$$(x,y)=(4,9) \ \text{and} \ (x,y)=(9,4) $$
as confirmed by the following graphical representation ((a) in blue and (b) in red) :
Best Answer
Since you require the sum to be zero, all you need is to compute the numerator of the LHS when put over a common denominator, since regardless of the value of the denominator--as long as it is nonzero--the LHS is zero only if the numerator is zero. Then once you solve for the roots of the numerator, check the validity of the solution set by substitution.