\( 4 \times \) Which equation has both 8 and -8 as possible values of \( y \) ? \( 4 \times \quad \) B \( y^{2}=-64 \) \( 4 \times \quad \) C \( y^{3}=\frac{1}{64} \) \( 4 \times \quad y^{3}=64 \)

We wish to determine which equation has both \(8\) and \(-8\) as solutions for \(y\). An equation that has both \(8\) and \(-8\) as solutions must be one that “loses” the sign information. A typical example is a square. In other words, an equation of the form \[ y^2 = k \] will have two solutions, namely \[ y = \sqrt{k}\quad \text{and} \quad y = -\sqrt{k}, \] provided \( k \ge 0 \). To have \(8\) and \(-8\) as solutions, we need: \[ \sqrt{k} = 8\quad \Longrightarrow\quad k = 8^2 = 64. \] Thus, the equation should be: \[ y^2 = 64. \] Now let’s check the given options: 1. Option B is \[ y^2 = -64. \] Since the square of any real number is nonnegative, there is no real number \(y\) for which \(y^2 = -64\). Hence, this option does not work. 2. Option C is \[ y^3 = \frac{1}{64}. \] Cubing is a one-to-one function (its inverse, the cube root, is also one-to-one). Therefore, there is only one real solution for \(y\). In fact, \[ y = \sqrt[3]{\tfrac{1}{64}} = \frac{1}{4}, \] which is not \(8\) or \(-8\). 3. The other provided option is \[ y^3 = 64. \] Again, because the cube function is one-to-one, it has exactly one real solution: \[ y = \sqrt[3]{64} = 4. \] Thus, this option does not give both \(8\) and \(-8\). None of the given equations have both \(8\) and \(-8\) as solutions. The correct equation that would have \(8\) and \(-8\) as solutions is: \[ y^2 = 64. \]