Certainly! Let's simplify the given expression step-by-step:
Given: [tex]\(\frac{27^{n+2} - 6 \cdot 3^{3n+3}}{3^n \times 9^{n+2}}\)[/tex]
### Step 1: Rewrite the bases in terms of powers of 3.
Notice that [tex]\(27\)[/tex] can be written as [tex]\(3^3\)[/tex] and [tex]\(9\)[/tex] can be written as [tex]\(3^2\)[/tex]. Substituting these, we get:
[tex]\[ 27^{n+2} = (3^3)^{n+2} = 3^{3(n+2)} \][/tex]
[tex]\[ 9^{n+2} = (3^2)^{n+2} = 3^{2(n+2)} \][/tex]
So the expression becomes:
[tex]\[ \frac{3^{3(n+2)} - 6 \cdot 3^{3n+3}}{3^n \times 3^{2(n+2)}} \][/tex]
### Step 2: Simplify the exponents.
First, simplify the exponents in the numerator:
[tex]\[ 3^{3(n+2)} = 3^{3n+6} \][/tex]
[tex]\[ 3^{3n+3} \][/tex]
And the denominator:
[tex]\[ 3^n \times 3^{2(n+2)} = 3^n \times 3^{2n+4} = 3^{n + 2n + 4} = 3^{3n+4} \][/tex]
Now, the expression is:
[tex]\[ \frac{3^{3n+6} - 6 \cdot 3^{3n+3}}{3^{3n+4}} \][/tex]
### Step 3: Divide each term in the numerator by the denominator.
We can split the fraction into two separate terms:
[tex]\[ \frac{3^{3n+6}}{3^{3n+4}} - \frac{6 \cdot 3^{3n+3}}{3^{3n+4}} \][/tex]
Simplify each term:
1. For the first term:
[tex]\[ \frac{3^{3n+6}}{3^{3n+4}} = 3^{(3n+6) - (3n+4)} = 3^2 = 9 \][/tex]
2. For the second term:
[tex]\[ \frac{6 \cdot 3^{3n+3}}{3^{3n+4}} = 6 \cdot 3^{(3n+3) - (3n+4)} = 6 \cdot 3^{-1} = 6 \cdot \frac{1}{3} = 2 \][/tex]
Putting these together, we get:
[tex]\[ 9 - 2 = 7 \][/tex]
Thus, the simplified value of the given expression is:
[tex]\[ \boxed{7} \][/tex]
