### 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, which means it might be difficult to solve directly by factoring or applying a formula.
However, I notice that the left-hand side is symmetric in x, meaning that if x is a solution, then so is 2-x.
This suggests that there might be some simplification possible by making a substitution.
Let me try letting y = x - 1, so that 2 - x = 1 - y. 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 $y^4+6y^2-16=0$.
Now this looks more manageable.
It is a quadratic equation in $y^2$, so I can try to factor it or use the quadratic formula.
Factoring seems easier, since I notice that 16 has only a few factors.
I try $(y^2+8)(y^2-2)=0$ and it works.
Then, either $y^2+8=0$ or $y^2-2=0$.
The first equation has no real solutions, since $y^2$ cannot be negative.
The second equation has two real solutions: $y=\pm\sqrt{2}$.
But remember, I defined y = x - 1, so I need to undo that substitution to find the solutions for x.
Adding 1 to both sides, I get $x=y+1=\pm\sqrt{2}+1$.
These are the only two real solutions to the original equation.