Answer :

Certainly! Let's simplify the given expression step by step:

Given:
[tex]\[ \left(\frac{3^n \times 9^{n+1}}{3^{n-1} \times 9^{n-1}}\right) \][/tex]

First, we recall that [tex]\(9\)[/tex] can be written as [tex]\(3^2\)[/tex]. Therefore, we can rewrite the expression with base [tex]\(3\)[/tex]:

[tex]\[ 9 = 3^2 \][/tex]

Substituting [tex]\(9\)[/tex] in the expression:

[tex]\[ \left(\frac{3^n \times (3^2)^{n+1}}{3^{n-1} \times (3^2)^{n-1}}\right) \][/tex]

Next, we simplify the exponents. Using the property of exponents [tex]\((a^m)^n = a^{mn}\)[/tex]:

[tex]\[ (3^2)^{n+1} = 3^{2(n+1)} = 3^{2n + 2} \][/tex]
[tex]\[ (3^2)^{n-1} = 3^{2(n-1)} = 3^{2n - 2} \][/tex]

Substituting these back into the expression, we get:

[tex]\[ \left(\frac{3^n \times 3^{2n + 2}}{3^{n-1} \times 3^{2n - 2}}\right) \][/tex]

Using the property of exponents [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we can combine the exponents in the numerator and the denominator:

In the numerator:
[tex]\[ 3^n \times 3^{2n + 2} = 3^{n + 2n + 2} = 3^{3n + 2} \][/tex]

In the denominator:
[tex]\[ 3^{n-1} \times 3^{2n-2} = 3^{(n-1) + (2n-2)} = 3^{n - 1 + 2n - 2} = 3^{3n - 3} \][/tex]

So, the expression simplifies to:

[tex]\[ \left(\frac{3^{3n + 2}}{3^{3n - 3}}\right) \][/tex]

Using the property [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex]:

[tex]\[ 3^{(3n + 2) - (3n - 3)} = 3^{3n + 2 - 3n + 3} = 3^{5} \][/tex]

Finally, calculating [tex]\(3^5\)[/tex]:

[tex]\[ 3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243 \][/tex]

Thus, the simplified expression is:

[tex]\[ \boxed{243} \][/tex]