To solve the equation \( x^8 = -8 \), we first note that \( x^8 \) represents the eighth power of \( x \), which is always non-negative for real numbers.
Therefore, there are no real solutions to the equation \( x^8 = -8 \).
However, if we consider complex solutions, we can rewrite the equation as:
\[ x^8 = -8 \]
We can express \(-8\) in polar form:
\[ -8 = 8(\cos(\pi) + i\sin(\pi)) \]
Now we can express this in exponential form:
\[ -8 = 8e^{i\pi} \]
Next, we can take the eighth root of both sides. The general formula for the \(n\)-th roots of a complex number is:
\[ r^{1/n} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left(\frac{\theta + 2k\pi}{n}\right) \right) \]
for \( k = 0, 1, 2, \ldots, n-1 \).
In our case, \( r = 8 \) and \( \theta = \pi \), and we want the eighth roots (\(n = 8\)):
\[ r^{1/8} = 8^{1/8} = 2 \]
Thus, the solution for \( x \) is:
\[ x_k = 2 \left( \cos\left(\frac{\pi + 2k\pi}{8}\right) + i \sin\left(\frac{\pi + 2k\pi}{8}\right) \right) \]
for \( k = 0, 1, 2, ..., 7 \).
Calculating each of these:
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For \( k = 0 \): \[ x_0 = 2 \left( \cos\left(\frac{\pi}{8}\right) + i \sin\left(\frac{\pi}{8}\right) \right) \]
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For \( k = 1 \): \[ x_1 = 2 \left( \cos\left(\frac{3\pi}{8}\right) + i \sin\left(\frac{3\pi}{8}\right) \right) \]
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For \( k = 2 \): \[ x_2 = 2 \left( \cos\left(\frac{5\pi}{8}\right) + i \sin\left(\frac{5\pi}{8}\right) \right) \]
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For \( k = 3 \): \[ x_3 = 2 \left( \cos\left(\frac{7\pi}{8}\right) + i \sin\left(\frac{7\pi}{8}\right) \right) \]
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For \( k = 4 \): \[ x_4 = 2 \left( \cos\left(\frac{9\pi}{8}\right) + i \sin\left(\frac{9\pi}{8}\right) \right) \]
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For \( k = 5 \): \[ x_5 = 2 \left( \cos\left(\frac{11\pi}{8}\right) + i \sin\left(\frac{11\pi}{8}\right) \right) \]
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For \( k = 6 \): \[ x_6 = 2 \left( \cos\left(\frac{13\pi}{8}\right) + i \sin\left(\frac{13\pi}{8}\right) \right) \]
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For \( k = 7 \): \[ x_7 = 2 \left( \cos\left(\frac{15\pi}{8}\right) + i \sin\left(\frac{15\pi}{8}\right) \right) \]
In summary, the eight solutions to the equation \( x^8 = -8 \) are given by:
\[ x_k = 2 \text{cis}\left(\frac{\pi + 2k\pi}{8}\right), \quad k = 0, 1, 2, 3, 4, 5, 6, 7 \]
where \( \text{cis}(\theta) \) is shorthand for \( \cos(\theta) + i \sin(\theta) \).