To simplify the expression \( \frac{32^3 \times 9^5}{8^3 \times 6^6} \), we can start by expressing each number in terms of its prime factors:

1. \( 32 = 2^5 \)
2. \( 9 = 3^2 \)
3. \( 8 = 2^3 \)
4. \( 6 = 2 \times 3 \)

Now, substitute these into the expression:

\[
32^3 = (2^5)^3 = 2^{15}
\]
\[
9^5 = (3^2)^5 = 3^{10}
\]
\[
8^3 = (2^3)^3 = 2^9
\]
\[
6^6 = (2 \times 3)^6 = 2^6 \times 3^6
\]

Substitute these back into the original expression:

\[
\frac{32^3 \times 9^5}{8^3 \times 6^6} = \frac{2^{15} \times 3^{10}}{2^9 \times 2^6 \times 3^6}
\]

Combine the exponents of the same bases in the denominator:

\[
2^9 \times 2^6 = 2^{9+6} = 2^{15}
\]

So the expression becomes:

\[
\frac{2^{15} \times 3^{10}}{2^{15} \times 3^6}
\]

Now, cancel out the common factors in the numerator and the denominator:

\[
\frac{2^{15}}{2^{15}} = 1
\]
\[
\frac{3^{10}}{3^6} = 3^{10-6} = 3^4
\]

Thus, the simplified expression is:

\[
3^4
\]

Finally, calculate \( 3^4 \):

\[
3^4 = 3 \times 3 \times 3 \times 3 = 81
\]

So, the simplified value of the expression is \( 81 \).