To solve the equation \( x^8 = -8 \), we first note that the left side of the equation, \( x^8 \), is always non-negative for any real number \( x \). This means that \( x^8 \) cannot equal a negative number, such as \(-8\).
Thus, there are no real solutions to the equation \( x^8 = -8 \).
However, if we allow for complex solutions, we can rewrite the equation in terms of complex numbers:
Taking the eighth root of both sides, we have:
\[ x = (-8)^{1/8} \]
To solve this in the complex plane, we can express \(-8\) in polar form. The complex number \(-8\) can be written as:
\[ -8 = 8 \text{cis} (180^\circ) = 8 \text{cis} \left(\pi\text{ (in radians)}\right) \]
Now, using De Moivre's Theorem:
\[ (-8)^{1/8} = 8^{1/8} \text{cis} \left(\frac{\pi + 2k\pi}{8}\right) \quad \text{for } k = 0, 1, 2, \ldots, 7 \]
Calculating \( 8^{1/8} \):
\[ 8^{1/8} = 2^{3/8} = 2^{3/8} \]
Thus, the solutions can be expressed as:
\[ x_k = 2^{3/8} \text{cis} \left(\frac{\pi + 2k\pi}{8}\right) \quad \text{for } k = 0, 1, 2, \ldots, 7 \]
For real solutions, there are none. If the problem only expects real solutions, then the answer would be:
No real solutions.
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