### Instruction: Find all real solutions to $x^4+(2-x)^4=34$.  Enter all the solutions, separated by commas.### Response: This is a quartic equation, but I notice that it is symmetrical around x = 1.
That is, if I replace x with 2 - x, I get the same equation.
So maybe I can use that to simplify the problem somehow.
If I let y = x - 1, then the equation becomes (y + 1)^4 + (1 - y)^4 = 34.
Expanding the binomials, I get y^4 + 4y^3 + 6y^2 + 4y + 1 + y^4 - 4y^3 + 6y^2 - 4y + 1 = 34.
Simplifying, I get 2y^4 + 12y^2 + 2 = 34, or y^4 + 6y^2 - 16 = 0.
This looks like a quadratic equation in y^2, so maybe I can factor it or use the quadratic formula.
Factoring, I get (y^2 + 8)(y^2 - 2) = 0.
So y^2 = -8 or y^2 = 2.
The first case has no real solutions, since y^2 can't be negative.
The second case gives y = +/- sqrt(2).
Now I have to go back to x, using x = y + 1.
So x = sqrt(2) + 1 or x = -sqrt(2) + 1.
These are the two real solutions to the original equation.