d) \( \frac{2^{a+1} \cdot 3^{a-1}}{6^{a}} \) g) \( \frac{2^{1+2 n} \cdot 8^{1-3 n}}{4^{n+1}} \) j) \( \frac{(x y z)^{o}}{2^{-3}} \)

We are given three expressions to simplify. Let’s work through each one step‐by‐step. ───────────────────────────── (d) Simplify  (2^(a+1) · 3^(a–1)) / (6^a) Step 1. Write 6^a as (2·3)^a = 2^a · 3^a. Step 2. Substitute into the original expression:   (2^(a+1) · 3^(a–1)) / (2^a · 3^a) Step 3. Simplify the powers of 2:   2^(a+1) / 2^a = 2^( (a+1) – a) = 2^1 = 2 Step 4. Simplify the powers of 3:   3^(a–1) / 3^a = 3^( (a–1) – a) = 3^(–1) = 1/3 Step 5. Multiply the simplified parts:   2 · (1/3) = 2/3 So, answer (d): 2/3 ───────────────────────────── (g) Simplify  (2^(1+2n) · 8^(1–3n)) / (4^(n+1)) Step 1. Express 8 and 4 as powers of 2:   8 = 2^3  and  4 = 2^2 Step 2. Write 8^(1–3n) as (2^3)^(1–3n) = 2^(3(1–3n)) = 2^(3 – 9n). Step 3. Write 4^(n+1) as (2^2)^(n+1) = 2^(2n+2). Step 4. Now substitute these back:   (2^(1+2n) · 2^(3–9n)) / 2^(2n+2) Step 5. Combine the exponents in the numerator:   2^( (1+2n)+(3–9n) ) = 2^(4 – 7n) Step 6. Now the expression is:   2^(4 – 7n) / 2^(2n+2) = 2^( (4 – 7n) – (2n + 2) ) Step 7. Simplify the exponent:   (4 – 7n) – (2n + 2) = 4 – 7n – 2n – 2 = 2 – 9n So, answer (g): 2^(2–9n) ───────────────────────────── (j) Simplify  ((x y z)^o) / (2^(–3)) Note: Here the exponent “o” is treated as a variable (not the number zero). Step 1. Recognize that dividing by 2^(–3) is the same as multiplying by 2^3:   ((x y z)^o) / (2^(–3)) = (x y z)^o · 2^3 Step 2. Since 2^3 = 8, we have:   8 · (x y z)^o So, answer (j): 8·(xyz)^o ───────────────────────────── Summary of Answers: (d) 2/3 (g) 2^(2–9n) (j) 8·(xyz)^o