### Instruction: Find all solutions $x$ (real and otherwise) to the equation
\[x^4+64=0.\]Enter all the solutions, separated by commas.### Response: To solve this equation, I want to factor it into simpler expressions that I can set equal to zero individually.
I notice that $x^4+64$ is a sum of two perfect squares, since $x^4=(x^2)^2$ and $64=8^2$.
This makes me think of using the identity $a^2+b^2=(a+bi)(a-bi)$, where $i$ is the imaginary unit that satisfies $i^2=-1$.
So, I can write $x^4+64=(x^2+8i)(x^2-8i)$.
Now, I can set each factor equal to zero and solve for $x$.
For the first factor, I get $x^2+8i=0$, which implies that $x^2=-8i$.
To find $x$, I need to take the square root of both sides, but I have to be careful with the sign and the imaginary part.
I recall that the square root of $i$ is $\frac{1}{\sqrt{2}}(1+i)$, so I can use that to simplify the square root of $-8i$.
I get $\sqrt{-8i}=\sqrt{-8}\sqrt{i}=\pm 2i\sqrt{2}\frac{1}{\sqrt{2}}(1+i)=\pm 2i(1+i)$.
Therefore, $x=\pm\sqrt{-8i}=\pm 2i(1+i)$.
For the second factor, I get $x^2-8i=0$, which implies that $x^2=8i$.
To find $x$, I need to take the square root of both sides again, but this time I have a positive imaginary part.
I recall that the square root of $-i$ is $\frac{1}{\sqrt{2}}(1-i)$, so I can use that to simplify the square root of $8i$.
I get $\sqrt{8i}=\sqrt{8}\sqrt{i}=\pm 2\sqrt{2}\frac{1}{\sqrt{2}}(1+i)=\pm 2(1+i)$.
Therefore, $x=\pm\sqrt{8i}=\pm 2(1+i)$.
To summarize, I have found four solutions for $x$: $2i(1+i)$, $-2i(1+i)$, $2(1+i)$, and $-2(1+i)$.