### Instruction: Find all solutions $x$ (real and otherwise) to the equation
\[x^4+64=0.\]Enter all the solutions, separated by commas.### Response: This is a quartic equation, so I could try to factor it into two quadratic factors.
To do that, I notice that $x^4+64$ is a sum of two perfect squares: $x^4=(x^2)^2$ and $64=8^2$.
There is a formula for factoring a sum of two perfect squares: $a^2+b^2=(a+bi)(a-bi)$, where $i$ is the imaginary unit such that $i^2=-1$.
Using this formula, I can write $x^4+64=(x^2+8i)(x^2-8i)$.
Now I have two quadratic factors, and I can use the quadratic formula to find their roots.
For the first factor, $x^2+8i=0$, I get $x=\pm\sqrt{-8i}$.
To simplify this, I can use the polar form of complex numbers: $-8i=8e^{i\frac{3\pi}{2}}$, since it has magnitude 8 and angle $\frac{3\pi}{2}$ radians in the complex plane.
Then, using the property that $\sqrt{e^{ix}}=e^{i\frac{x}{2}}$, I get $\sqrt{-8i}=\pm 2\sqrt{2}e^{i\frac{3\pi}{4}}$.
This means that $x=\pm 2\sqrt{2}e^{i\frac{3\pi}{4}}$ are two solutions.
For the second factor, $x^2-8i=0$, I get $x=\pm\sqrt{8i}$.
Using the polar form again, I get $8i=8e^{i\frac{\pi}{2}}$, since it has magnitude 8 and angle $\frac{\pi}{2}$ radians in the complex plane.
Then, using the same property as before, I get $\sqrt{8i}=\pm 2\sqrt{2}e^{i\frac{\pi}{4}}$.
This means that $x=\pm 2\sqrt{2}e^{i\frac{\pi}{4}}$ are the other two solutions.
Therefore, the equation has four solutions, and they are $x=\pm 2\sqrt{2}e^{i\frac{\pi}{4}}, \pm 2\sqrt{2}e^{i\frac{3\pi}{4}}$.