Problem
Given that \( x = \sqrt[3]{27} \) and \( y = \cos(\frac{\pi}{3}) \), solve for \( z = 3x^2 - 2y \).
Solution
Step 1 ：First, evaluate the cube root: \( x = \sqrt[3]{27} = 3 \).
Step 2 ：Next, evaluate the cosine function: \( y = \cos(\frac{\pi}{3}) = \frac{1}{2} \).
Step 3 ：Finally, substitute \( x \) and \( y \) into the equation: \( z = 3x^2 - 2y = 3(3)^2 - 2(\frac{1}{2}) = 27 - 1 = 26 \).
