To simplify the expression \((7^{-3})^7 \cdot 7^{20}/6^{-2}\) and ensure we only have positive exponents, we will apply the properties of integer exponents.
Step 1: Simplify \((7^{-3})^7\)
Using the power of a power property, which states that \((a^m)^n = a^{m \cdot n}\):
\[ (7^{-3})^7 = 7^{-3 \cdot 7} = 7^{-21} \]
Step 2: Combine with \(7^{20}\)
Now we rewrite our expression:
\[ 7^{-21} \cdot 7^{20} \]
Using the product of powers property, which states that \(a^m \cdot a^n = a^{m+n}\):
\[ 7^{-21 + 20} = 7^{-1} \]
Step 3: Simplify the overall expression including division by \(6^{-2}\)
Now we have:
\[ \frac{7^{-1}}{6^{-2}} \]
Using the property of negative exponents, which states that \(a^{-m} = \frac{1}{a^m}\), we can rewrite this:
\[ \frac{7^{-1}}{6^{-2}} = 7^{-1} \cdot 6^{2} = \frac{6^2}{7^{1}} = \frac{36}{7} \]
Final Answer
The expression \((7^{-3})^7 \cdot 7^{20}/6^{-2}\) simplifies to:
\[ \frac{36}{7} \]
Thus, the answer is \(\frac{36}{7}\).