To solve the equation \( y^3 = 64 \), you need to find the cube roots of 64.
The cube root of 64 is 4, since \( 4^3 = 64 \).
Thus, the real solution is:
- \( y = 4 \)
However, cube roots also have complex solutions. The general formula for the cube roots of a number can be expressed as:
\[ y = \sqrt[3]{r} \text{cis} \left( \frac{\theta + 2k\pi}{3} \right) \quad \text{for } k = 0, 1, 2 \]
For \( 64 = 64 \text{cis}(0) \):
- For \( k = 0 \): \( y_0 = 4 \)
- For \( k = 1 \): \( y_1 = 4 \text{cis} \left( \frac{2\pi}{3} \right) = 4 \left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} \right) = -2 + 2i\sqrt{3} \)
- For \( k = 2 \): \( y_2 = 4 \text{cis} \left( \frac{4\pi}{3} \right) = 4 \left( -\frac{1}{2} - i\frac{\sqrt{3}}{2} \right) = -2 - 2i\sqrt{3} \)
In this context, the only real solution is \( y = 4 \). The other values \( -8, 8, -4 \) do not satisfy the equation \( y^3 = 64 \):
- \( (-8)^3 = -512 \)
- \( (8)^3 = 512 \)
- \( (-4)^3 = -64 \)
Thus, the only correct answer from the options given is:
D: 4