To solve the equation \( x^8 = -8 \), we first note that the left side, \( x^8 \), represents the fourth power of \( x^2 \) and is always non-negative for real numbers. This means that there is no real solution since \( -8 \) is negative.
However, if we were to consider complex solutions, we can proceed as follows:
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Rewrite the equation:
\[ x^8 = -8 \]
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To express \(-8\) in polar form, we recognize that:
\[ -8 = 8(\cos(\pi) + i \sin(\pi)) \]
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We can find the eighth roots of \(-8\) using De Moivre's Theorem. The magnitude \( r \) is \( 8 \), and the angle \( \theta = \pi + 2k\pi \) (where \( k \) is any integer for multiple roots). The eighth roots are given by:
\[ r^{1/8} = 8^{1/8} = 2 \]
and
\[ \theta_n = \frac{\pi + 2k\pi}{8} \quad (k = 0, 1, 2, \ldots, 7) \]
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Thus, the angles will be:
- For \( k=0 \): \( \theta_0 = \frac{\pi}{8} \)
- For \( k=1 \): \( \theta_1 = \frac{\pi}{8} + \frac{2\pi}{8} = \frac{3\pi}{8} \)
- For \( k=2 \): \( \theta_2 = \frac{\pi}{8} + \frac{4\pi}{8} = \frac{5\pi}{8} \)
- For \( k=3 \): \( \theta_3 = \frac{\pi}{8} + \frac{6\pi}{8} = \frac{7\pi}{8} \)
- For \( k=4 \): \( \theta_4 = \frac{\pi}{8} + \frac{8\pi}{8} = \frac{9\pi}{8} \)
- For \( k=5 \): \( \theta_5 = \frac{\pi}{8} + \frac{10\pi}{8} = \frac{11\pi}{8} \)
- For \( k=6 \): \( \theta_6 = \frac{\pi}{8} + \frac{12\pi}{8} = \frac{13\pi}{8} \)
- For \( k=7 \): \( \theta_7 = \frac{\pi}{8} + \frac{14\pi}{8} = \frac{15\pi}{8} \)
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Therefore, the roots are:
\[ x_k = 2 \left( \cos(\theta_k) + i \sin(\theta_k) \right) \]
Since the question seems to be looking for real solutions, the only real number that satisfies \( x^8 = -8 \) is that there are none.
None of the provided responses (\( x=0, x=-1, x=-64, x=64 \)) are solutions for the equation \( x^8 = -8 \). Thus, the answer is:
There are no real solutions.