If $$\frac{(9^n \times 3^2 \times 3^n – (27)^n )}{((3^3 )^5 \times 2^3)}$$​= $$\frac{1}{27}$$​then the value of 'n' is

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Correct option is A

​$$\text{Given,} \quad \frac{9^n \times 3^2 \times 3^n - 27^n}{(3^3)^5 \times 2^3} = \frac{1}{27}\text{Simplifying the numerator:} \quad 9^n \times 3^2 \times 3^n - 27^n = (9 \times 3)^n \times 9 - 27^n = 27^n \times 9 - 27^n = 27^n (9 - 1) = 8 \times 27^n\text{Simplifying the denominator:} \quad (3^3)^5 \times 2^3 = 27^5 \times 8\text{Now, the fraction becomes:} \quad \frac{8 \times 27^n}{8 \times 27^5} = 27^{n - 5}\text{Equating to} \quad \frac{1}{27} = 27^{-1}, \quad \text{we get:} \quad 27^{n - 5} = 27^{-1}\text{Therefore,} \quad n - 5 = -1 \quad \Longrightarrow \quad n = 4$$​​