### 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, so it might be hard to solve by factoring or using the rational root theorem.
I notice that the equation is symmetric in $x$ and $2-x$, which means that if $x$ is a solution, so is $2-x$.
This suggests that I can try to make a substitution, such as $u=x-1$, to simplify the equation.
If I do that, I get $(u+1)^4+(1-u)^4=34$, which expands to $2u^4+12u^2+2=34$.
Dividing by 2, I get $u^4+6u^2-16=0$.
This looks like a quadratic equation in $u^2$, so I can try to factor it as $(u^2+a)(u^2+b)=0$, where $ab=-16$ and $a+b=6$.
I see that $a=8$ and $b=-2$ work, so I get $(u^2+8)(u^2-2)=0$.
This means that $u^2=-8$ or $u^2=2$.
The first case has no real solutions, since $u^2$ cannot be negative.
The second case has two real solutions, $u=\pm\sqrt{2}$.
To find the solutions for $x$, I need to undo the substitution, so I get $x=u+1=\pm\sqrt{2}+1$.
Therefore, the only two real solutions to the original equation are $x=\sqrt{2}+1$ and $x=-\sqrt{2}+1$.