The given expression is:

$\frac{3^{-10} \cdot 9^8}{(-3)^2} - 9^{-6} \cdot 27^5$

Let’s break this down step by step to solve it.

Step 1: Simplifying each part

1. $3^{-10} \cdot 9^8$

We know that $9 = 3^2$, so:

$9^8 = (3^2)^8 = 3^{16}$

Now, the first part of the numerator becomes:

$3^{-10} \cdot 3^{16} = 3^{(-10 + 16)} = 3^6$

2. $(-3)^2$

This is simply:

$(-3)^2 = 9$

So, the first term is:

$\frac{3^6}{9}$

Since $9 = 3^2$, we can simplify further:

$\frac{3^6}{3^2} = 3^{6-2} = 3^4$

3. $9^{-6} \cdot 27^5$

We know that $9 = 3^2$ and $27 = 3^3$, so:

$9^{-6} = (3^2)^{-6} = 3^{-12}$ $27^5 = (3^3)^5 = 3^{15}$

Thus, the second term becomes:

$3^{-12} \cdot 3^{15} = 3^{(-12 + 15)} = 3^3$

Step 2: Combining the terms

Now, we have:

$3^4 - 3^3$

Factoring this expression gives:

$3^3(3 - 1) = 3^3 \cdot 2 = 27 \cdot 2 = 54$

Final Answer:

The value of the expression is $\boxed{54}$.

Would you like more details on any step or have any questions?

Here are 5 related questions:

1. How do we handle negative exponents like $3^{-10}$?
2. Can you explain how to simplify expressions with powers of the same base?
3. What are the rules for multiplying and dividing exponential expressions?
4. Why does $(-3)^2 = 9$ and not $-9$?
5. How can we factor expressions like $3^4 - 3^3$?

Tip: Always check for ways to rewrite numbers in terms of prime factors when simplifying exponents!