### Instruction: Find all the solutions to
\[\frac{1}{x^2 + 11x - 8} + \frac{1}{x^2 + 2x - 8} + \frac{1}{x^2 - 13x - 8} = 0.\]Enter all the solutions, separated by commas.### Response: I notice that the three fractions have the same denominator, except for the coefficient of x.
Maybe I can factor the denominators and see if there is a common factor.
I try to find two numbers that multiply to -8 and add to 11, 2, and -13 respectively.
I see that 8 and -1 work for the first fraction, so the denominator is (x + 8)(x - 1).
I see that 4 and -2 work for the second fraction, so the denominator is (x + 4)(x - 2).
I see that -8 and 1 work for the third fraction, so the denominator is (x - 8)(x + 1).
Now I can write the equation as
\[\frac{1}{(x + 8)(x - 1)} + \frac{1}{(x + 4)(x - 2)} + \frac{1}{(x - 8)(x + 1)} = 0.\]
I want to get rid of the fractions, so I multiply both sides by the product of the three denominators.
This gives me
\[(x + 4)(x - 2)(x - 8)(x + 1) + (x + 8)(x - 1)(x - 8)(x + 1) + (x + 8)(x - 1)(x + 4)(x - 2) = 0.\]
I notice that this is a sum of three products of four factors each, and each factor appears twice in the sum.
This means that I can factor out each factor and get a simpler expression.
I factor out (x + 8) and get
\[(x + 8)((x - 1)(x - 2)(x - 8) + (x - 1)(x + 1)(x - 8) + (x - 1)(x + 4)(x - 2)) = 0.\]
I factor out (x - 1) and get
\[(x + 8)(x - 1)((x - 2)(x - 8) + (x + 1)(x - 8) + (x + 4)(x - 2)) = 0.\]
I factor out (x - 8) and get
\[(x + 8)(x - 1)(x - 8)((x - 2) + (x + 1) + (x + 4)) = 0.\]
I simplify the last factor and get
\[(x + 8)(x - 1)(x - 8)(3x + 3) = 0.\]
Now I have a product of four factors equal to zero, so I can use the zero product property and set each factor equal to zero.
I get x = -8, x = 1, x = 8, and x = -1.
These are the four solutions to the equation.