### Instruction: Solve
\[(x - 3)^4 + (x - 5)^4 = -8.\]Enter all the solutions, separated by commas.### Response: This problem looks hard, but maybe I can simplify it by making a substitution.
Let me try letting y = x - 4, then x = y + 4, and plug that into the equation.
I get (y + 4 - 3)^4 + (y + 4 - 5)^4 = -8, which simplifies to (y + 1)^4 + (y - 1)^4 = -8.
This looks a little better, because now I have two terms that are almost the same, except for the sign of 1.
I wonder if I can use some algebraic identity to rewrite this expression.
I recall that (a + b)^2 = a^2 + 2ab + b^2 and (a - b)^2 = a^2 - 2ab + b^2, so maybe I can use that to expand the fourth powers.
I get (y + 1)^4 = [(y + 1)^2]^2 = (y^2 + 2y + 1)^2 and (y - 1)^4 = [(y - 1)^2]^2 = (y^2 - 2y + 1)^2.
Now I can substitute these into the equation and get (y^2 + 2y + 1)^2 + (y^2 - 2y + 1)^2 = -8.
I can expand the squares again and get y^4 + 4y^3 + 6y^2 + 4y + 1 + y^4 - 4y^3 + 6y^2 - 4y + 1 = -8.
I can simplify this by combining like terms and get 2y^4 + 12y^2 + 2 = -8.
I can subtract 2 from both sides and get 2y^4 + 12y^2 = -10.
I can divide both sides by 2 and get y^4 + 6y^2 = -5.
This looks like a quadratic equation in y^2, so I can try to factor it.
I get (y^2 + 5)(y^2 + 1) = 0.
This means that either y^2 + 5 = 0 or y^2 + 1 = 0.
I can solve for y by taking the square root of both sides.
I get y = ± i√5 or y = ± i, where i is the imaginary unit that satisfies i^2 = -1.
I can plug these values of y back into the original substitution and get x = y + 4.
I get x = 4 ± i√5 or x = 4 ± i.
These are the four complex solutions to the equation.