### Instruction: Find all solutions $x$ (real and otherwise) to the equation
\[x^4+64=0.\]Enter all the solutions, separated by commas.### Response: I know that $x^4$ is always non-negative for any real $x$, so adding 64 to it will make it always positive.
Therefore, there are no real solutions to this equation.
To find the non-real solutions, I will try to factor the expression $x^4+64$ into two quadratic factors.
I recall that the difference of two squares can be factored as $(a^2-b^2)=(a+b)(a-b)$.
I wonder if I can rewrite $x^4+64$ as a difference of two squares by adding and subtracting some term.
I notice that $x^4+64=(x^4+16x^2+64)-16x^2$, which is a difference of two squares: $(x^2+8)^2-(4x)^2$.
Now I can apply the formula and get $(x^2+8)^2-(4x)^2=(x^2+8+4x)(x^2+8-4x)$.
These are two quadratic factors, but they are not easily factorable into linear factors.
I will use the quadratic formula to find their roots.
The roots of $x^2+8+4x$ are $x=\frac{-4\pm\sqrt{16-4(1)(8)}}{2(1)}=\frac{-4\pm\sqrt{-16}}{2}=\frac{-4\pm 4i}{2}=-2\pm 2i$.
The roots of $x^2+8-4x$ are $x=\frac{4\pm\sqrt{16-4(1)(8)}}{2(1)}=\frac{4\pm\sqrt{-16}}{2}=\frac{4\pm 4i}{2}=2\pm 2i$.
Therefore, the four solutions to the equation are $x=-2\pm 2i$ and $x=2\pm 2i$.