Simplify and write in exponential form:
 \[\dfrac{{{{\left( {{3^2}} \right)}^3} \times {{\left( { - 2} \right)}^5}}}{{{{\left( { - 2} \right)}^3}}}\]

Hint: In the given question, we have been given an expression which needs to be simplified. It can be easily done if we know the formula of division of two numbers with equal base but unequal exponent.
Formula Used:
For this question, we are going to use the formula for division of two numbers with equal base but unequal exponent, which is,
 \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]

Complete step-by-step answer:
The given expression is
 \[A = \dfrac{{{{\left( {{3^2}} \right)}^3} \times {{\left( { - 2} \right)}^5}}}{{{{\left( { - 2} \right)}^3}}}\]
To solve this, we just need to solve the denominator with its similar term in the numerator,
 \[A = \dfrac{{{{\left( {{3^2}} \right)}^3} \times {{\left( { - 2} \right)}^5}}}{{{{\left( { - 2} \right)}^3}}} = {\left( {{3^2}} \right)^3} \times {\left( { - 2} \right)^{5 - 3}}\]
But, we know, \[{\left( {{a^n}} \right)^m} = {a^{m \times n}}\] and \[{\left( { - a} \right)^{2m}} = {\left( a \right)^{2m}}\]
Hence, \[A = {\left( {{3^2}} \right)^3} \times {\left( { - 2} \right)^{5 - 3}} = {\left( 3 \right)^{2 \times 3}} \times {\left( { - 2} \right)^2} = {\left( 3 \right)^6} \times {\left( 2 \right)^2}\]
So, \[\dfrac{{{{\left( {{3^2}} \right)}^3} \times {{\left( { - 2} \right)}^5}}}{{{{\left( { - 2} \right)}^3}}} = {\left( 3 \right)^6} \times {\left( 2 \right)^2}\]
So, the correct answer is “ $ {\left( 3 \right)^6} \times {\left( 2 \right)^2} $ ”.

Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the concept or formula which contains the known and the unknown and pick the one which is the most suitable and the most effective for finding the answer of the given question. Then we use the results or finding of the concept and apply it to our question. It is really important to know and follow all the results of the concepts if we have to solve the question correctly, as one slightest error gives the incorrect result.