\( 4 \times \) Which equation has both 8 and -8 as possible values of \( y \) ? \( 4 \times \quad \) B \( y^{2}=-64 \) \( 4 \times \quad \) C \( y^{3}=\frac{1}{64} \) \( 4 \times \quad y^{3}=64 \)
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The Deep Dive
To determine which equation has both 8 and -8 as possible values of \( y \), we can first analyze the option \( B \) which is \( y^{2} = -64 \). Since the square of any real number cannot be negative, this equation cannot have real solutions. For option \( C \), \( y^{3} = \frac{1}{64} \) and solving for \( y \) gives \( y = \frac{1}{4} \), which does not include 8 or -8. Lastly, in the equation \( y^{3} = 64 \), taking the cube root gives us \( y = 4 \), which again does not include 8 or -8. Upon further inspection, none of the equations provided allow both 8 and -8 as solutions. In fact, neither option meets the criteria set by the problem statement, making them all unsuitable. Keep in mind, only equations where \( y^2 = k \) (where \( k > 0 \)) would include both positive and negative solutions, but finally, the equations listed do not comply!
