\( \begin{array}{ll}\text { a } 3^{2} \cdot 2=18 & \text { d } 4 \cdot 2^{3}-2^{3} \\ \text { b } \frac{4^{3}}{4} & \text { e } \frac{3^{3}+5}{2^{5}} \\ \text { c }(6-2)^{3} & \text { f } \frac{2^{4}+2^{3}}{2 \cdot 2^{2}}\end{array} \)

Did you know that the expression \( 4 \cdot 2^{3} - 2^{3} \) can be factored? This clever factorization transforms the challenge into a simpler one: \( (4-1) \cdot 2^{3} \). Simplifying this gives us \( 3 \cdot 8 = 24 \)! Also, when evaluating \( \frac{3^{3}+5}{2^{5}} \), keep in mind that \( 3^{3} \) equals 27. So, the expression turns into \( \frac{27 + 5}{32} \), leading to an answer of \( \frac{32}{32} = 1 \). Math can be a little puzzle, but it's fun to see how everything clicks together!