The expression you provided is:

$\frac{54^n - 18^{n-1} \cdot 3^{n+1}}{(3^n - 1)^2 \cdot 6^n}$

Let me break down the expression and simplify it step by step:

Step 1: Rewrite powers of numbers as products of primes

1. $54 = 2 \cdot 3^3$
2. $18 = 2 \cdot 3^2$
3. $6 = 2 \cdot 3$

Using these, we can express everything in terms of powers of 2 and 3:

$54^n = (2 \cdot 3^3)^n = 2^n \cdot 3^{3n}$ $18^{n-1} = (2 \cdot 3^2)^{n-1} = 2^{n-1} \cdot 3^{2(n-1)} = 2^{n-1} \cdot 3^{2n-2}$ $3^{n+1} \text{ remains the same.}$ $6^n = (2 \cdot 3)^n = 2^n \cdot 3^n$

Step 2: Substitute the rewritten terms

Now, let's substitute these into the original expression:

$\frac{2^n \cdot 3^{3n} - (2^{n-1} \cdot 3^{2n-2}) \cdot 3^{n+1}}{(3^n - 1)^2 \cdot 2^n \cdot 3^n}$

Step 3: Simplify the numerator

The second term in the numerator simplifies as follows:

$(2^{n-1} \cdot 3^{2n-2}) \cdot 3^{n+1} = 2^{n-1} \cdot 3^{2n-2 + n+1} = 2^{n-1} \cdot 3^{3n-1}$

Thus, the numerator becomes:

$2^n \cdot 3^{3n} - 2^{n-1} \cdot 3^{3n-1}$

Factor out $2^{n-1} \cdot 3^{3n-1}$:

$2^{n-1} \cdot 3^{3n-1} \cdot \left( 2 \cdot 3 - 1 \right)$

Since $2 \cdot 3 - 1 = 5$, the numerator becomes:

$5 \cdot 2^{n-1} \cdot 3^{3n-1}$

Step 4: Simplify the entire expression

Now, the whole expression becomes:

$\frac{5 \cdot 2^{n-1} \cdot 3^{3n-1}}{(3^n - 1)^2 \cdot 2^n \cdot 3^n}$

Cancel $2^{n-1}$ with $2^n$, and $3^n$ with part of the $3^{3n-1}$:

$\frac{5 \cdot 3^{2n-1}}{(3^n - 1)^2 \cdot 2}$

Thus, the simplified expression is:

$\boxed{\frac{5 \cdot 3^{2n-1}}{2 \cdot (3^n - 1)^2}}$

Let me know if you'd like further clarifications or details!

Here are five related questions to deepen understanding:

1. How do you simplify expressions involving exponents with different bases?
2. What are some strategies to factor more complex algebraic expressions?
3. How do powers of numbers behave under multiplication and division?
4. What are the general rules for simplifying fractions with powers?
5. How does one handle higher degree polynomials like $(3^n - 1)^2$?

Tip: Always try to factor common terms in both the numerator and denominator to simplify expressions efficiently!