### Instruction: Solve
\[(x - 3)^4 + (x - 5)^4 = -8.\]Enter all the solutions, separated by commas.### Response: This problem looks like it might involve complex numbers, since the sum of two squares is negative.
I also notice that the left-hand side is symmetric about x = 4, so that might help simplify the algebra.
Let me start by making a substitution: let u = x - 4, so that x = u + 4.
Then the equation becomes \[(u + 1)^4 + (u - 1)^4 = -8.\]
Now I can expand the binomials using the formula \[(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.\]
This gives me \[u^4 + 4u^3 + 6u^2 + 4u + 1 + u^4 - 4u^3 + 6u^2 - 4u + 1 = -8.\]
Simplifying, I get \[2u^4 + 12u^2 + 2 = -8.\]
Dividing everything by 2, I get \[u^4 + 6u^2 + 1 = -4.\]
Now I can solve for u^2 by moving everything to one side and factoring: \[u^4 + 6u^2 + 5 = 0.\]
\[(u^2 + 5)(u^2 + 1) = 0.\]
This means that either u^2 = -5 or u^2 = -1.
To find u, I need to take the square root of both sides, remembering to include both the positive and negative roots.
This gives me \[u = \pm \sqrt{-5} \quad \text{or} \quad u = \pm \sqrt{-1}.\]
Now I can use the fact that \[\sqrt{-a} = i\sqrt{a}\] where i is the imaginary unit that satisfies \[i^2 = -1.\]
This means that \[u = \pm i\sqrt{5} \quad \text{or} \quad u = \pm i.\]
To find x, I need to undo the substitution and add 4 to both sides.
This gives me \[x = u + 4 = \pm i\sqrt{5} + 4 \quad \text{or} \quad x = u + 4 = \pm i + 4.\]
These are the four solutions for x.