### 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 it's hard to imagine how two fourth powers can add up to a negative number.
Let me try to simplify it a bit by letting y = x - 4, which is the midpoint of 3 and 5.
Then the equation becomes \[(y + 1)^4 + (y - 1)^4 = -8.\]
I can expand the binomials using the formula \[(a \pm b)^4 = a^4 \pm 4a^3b + 6a^2b^2 \pm 4ab^3 + b^4.\]
Applying this to both terms, I get \[y^4 + 4y^3 + 6y^2 + 4y + 1 + y^4 - 4y^3 + 6y^2 - 4y + 1 = -8.\]
Simplifying, I get \[2y^4 + 12y^2 + 2 = -8.\]
Dividing by 2, I get \[y^4 + 6y^2 + 1 = -4.\]
Adding 4 to both sides, I get \[y^4 + 6y^2 + 5 = 0.\]
This looks like a quadratic equation in y^2, so I can try to factor it.
I look for two numbers that multiply to 5 and add to 6.
I see that 5 and 1 work, so I can write \[(y^2 + 5)(y^2 + 1) = 0.\]
Now I can use the zero product property to find the values of y^2 that make this true.
I get \[y^2 = -5 \quad \text{or} \quad y^2 = -1.\]
Taking the square root of both sides, I get \[y = \pm i\sqrt{5} \quad \text{or} \quad y = \pm i.\]
Remembering that y = x - 4, I can add 4 to both sides to find the values of x.
I get \[x = 4 \pm i\sqrt{5} \quad \text{or} \quad x = 4 \pm i.\]
These are the four solutions to the equation.