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**Hint:**In the given question, we are required to find the value of inverse of the square of the expression $\left( {3{x^2}{y^3}} \right)$. Given is a bracket with an expression involving two variables, x and y. So, we have to evaluate the inverse of the square of the term. Square is nothing but multiplying the same number with itself. So we will multiply the bracket with itself. Then each individual term in the first bracket is multiplied with that in the second term. Then if needed any mathematical operations, those will be performed. Or else we can use the standard and important identities used for expansion. Like those used in squaring or cubing. Those are the algebraic identities that come in significant use when solving such questions.

**Complete step by step answer:**

Given that: ${\left( {3{x^2}{y^3}} \right)^{ - 2}}$. So, we have to find the inverse of the square of the expression $\left( {3{x^2}{y^3}} \right)$.

Firstly, We have to evaluate the square of the expression and then find the inverse of the obtained result. So, we get,

${\left( {3{x^2}{y^3}} \right)^2} = \left( {9{x^4}{y^6}} \right)$

Now, we have to find the inverse of the obtained expression to get the required final result.Inverse of an expression is reciprocal of the expression.So, Inverse of $\left( {9{x^4}{y^6}} \right)$ is $\dfrac{1}{{9{x^4}{y^6}}}$.

**So, the expression ${\left( {3{x^2}{y^3}} \right)^{ - 2}}$ given to us can be simplified as $\dfrac{1}{{9{x^4}{y^6}}}$.**

**Note:**In this problem we are taking the inverse of square of a term. After that when we are simplifying the product we have added the terms with the same coefficient. If we are asked to find the cube of the term then we take the same and multiply the term three times. There are various important identities that help us in finding the square and cube expansion of the same problem directly.

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