a) \( \frac{63^{1-n} \cdot 7^{n-1} \cdot 81^{-1}}{4^{n+2} \cdot 36^{-n-1}} \)

To simplify the expression \( \frac{63^{1-n} \cdot 7^{n-1} \cdot 81^{-1}}{4^{n+2} \cdot 36^{-n-1}} \), we first express each base in terms of its prime factors: - \( 63 = 7^2 \cdot 3^2 \) - \( 81 = 3^4 \) - \( 4 = 2^2 \) - \( 36 = 6^2 = 2^2 \cdot 3^2 \) Now we can rewrite the expression: \[ \frac{(7^2 \cdot 3^2)^{1-n} \cdot 7^{n-1} \cdot (3^4)^{-1}}{(2^2)^{n+2} \cdot (2^2 \cdot 3^2)^{-(n+1)}} \] Next, simplify each component: **Numerator:** \[ (7^2)^{1-n} = 7^{2(1-n)} = 7^{2 - 2n} \] \[ (3^2)^{1-n} = 3^{2(1-n)} = 3^{2 - 2n} \] \[ 7^{n-1} = 7^{n-1} \] \[ (3^4)^{-1} = 3^{-4} \] Thus, the numerator becomes: \[ 7^{2-2n} \cdot 3^{2-2n} \cdot 7^{n-1} \cdot 3^{-4} = 7^{(2-2n)+(n-1)} \cdot 3^{(2-2n)-4} = 7^{n+1-2n} \cdot 3^{-2n-2} = 7^{1-n} \cdot 3^{-2n-2} \] **Denominator:** \[ (2^2)^{n+2} = 2^{2(n+2)} = 2^{2n+4} \] For \( 36^{-(n+1)} \): \[ (2^2 \cdot 3^2)^{-(n+1)} = (2^2)^{-(n+1)} \cdot (3^2)^{-(n+1)} = 2^{-2(n+1)} \cdot 3^{-2(n+1)} = 2^{-2n - 2} \cdot 3^{-2n - 2} \] Putting it all together for the denominator: \[ 2^{2n + 4} \cdot (2^{-2n - 2} \cdot 3^{-2n - 2}) = 2^{(2n + 4) - (2n + 2)} \cdot 3^{-2n - 2} = 2^{2} \cdot 3^{-2n - 2} = 4 \cdot 3^{-2n - 2} \] Finally, we combine the numerator and the denominator: \[ \frac{7^{1-n} \cdot 3^{-2n-2}}{4 \cdot 3^{-2n-2}} = \frac{7^{1-n}}{4} \] Thus, the simplified expression is: \[ \frac{7^{1-n}}{4} \]