### Instruction: Find all solutions $x$ (real and otherwise) to the equation
\[x^4+64=0.\]Enter all the solutions, separated by commas.### Response: This problem is asking me to find the roots of a quartic equation, which are the values of $x$ that make the equation true.
One possible strategy is to try to factor the equation into simpler expressions that can be solved more easily.
To do that, I need to look for common factors or patterns that can help me break down the quartic term.
I notice that $x^4+64$ is a sum of two perfect squares, since $x^4=(x^2)^2$ and $64=8^2$.
This suggests that I can use the identity $a^2+b^2=(a+bi)(a-bi)$, where $i$ is the imaginary unit that satisfies $i^2=-1$.
Applying this identity to $x^4+64$, I get $(x^2+8i)(x^2-8i)$.
Now I have a product of two quadratic factors that equals zero, so I can use the zero product property to set each factor equal to zero and solve for $x$.
For the first factor, I have $x^2+8i=0$, which implies that $x^2=-8i$.
Taking the square root of both sides, I get $x=\pm\sqrt{-8i}$.
To simplify this expression, I can use the polar form of complex numbers, where $-8i$ can be written as $8e^{-i\frac{\pi}{2}}$, since it has a magnitude of 8 and an angle of $-\frac{\pi}{2}$ radians with the positive real axis.
Then, by applying the rule $\sqrt{re^{i\theta}}=\sqrt{r}e^{i\frac{\theta}{2}}$, I get $x=\pm\sqrt{8}e^{-i\frac{\pi}{4}}$.
Expanding this using Euler's formula, I get $x=\pm\sqrt{8}(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))$.
Simplifying the trigonometric functions, I get $x=\pm\sqrt{8}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})$.
For the second factor, I have $x^2-8i=0$, which implies that $x^2=8i$.
Taking the square root of both sides, I get $x=\pm\sqrt{8i}$.
To simplify this expression, I can use the polar form of complex numbers, where $8i$ can be written as $8e^{i\frac{\pi}{2}}$, since it has a magnitude of 8 and an angle of $\frac{\pi}{2}$ radians with the positive real axis.
Then, by applying the rule $\sqrt{re^{i\theta}}=\sqrt{r}e^{i\frac{\theta}{2}}$, I get $x=\pm\sqrt{8}e^{i\frac{\pi}{4}}$.
Expanding this using Euler's formula, I get $x=\pm\sqrt{8}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$.
Simplifying the trigonometric functions, I get $x=\pm\sqrt{8}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})$.
Therefore, the four solutions for $x$ are $\sqrt{8}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})$, $-\sqrt{8}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})$, $\sqrt{8}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})$, and $-\sqrt{8}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})$.