Write the expanded form of${{\left[ \dfrac{3}{2}x+1 \right]}^{3}}$ .

Hint: Here, in this given question, we can use the formula for the expansion of cube of the sum of two numbers; if a and b are two numbers, then\[{{\left( a+b \right)}^{3}}={{\left( a \right)}^{3}}+{{\left( b \right)}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}={{\left( a \right)}^{3}}+{{\left( b \right)}^{3}}+3ab\left( a+b \right)\] , to solve the given question. We may substitute a and b as $\dfrac{3}{2}x$ and 1 respectively in the formula and then simplify it accordingly in order to find our answer.

Complete step by step solution:
In this given question, we are asked to write the expanded form of${{\left[ \dfrac{3}{2}x+1 \right]}^{3}}$ .
Now, we know that the formula for the expansion of cube of the sum of two numbers is:
Let a and b be two numbers, then the formula for the cube of their sum is
\[{{\left( a+b \right)}^{3}}={{\left( a \right)}^{3}}+{{\left( b \right)}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}={{\left( a \right)}^{3}}+{{\left( b \right)}^{3}}+3ab\left( a+b \right).........................(1.1)\]
Here, let us take a as $\dfrac{3}{2}x$ and b as 1.
So, by using equation 1.1, we get
 $\begin{align}
  & {{\left[ \dfrac{3}{2}x+1 \right]}^{3}}={{\left( \dfrac{3}{2}x \right)}^{3}}+{{\left( 1 \right)}^{3}}+3\times {{\left( \dfrac{3}{2}x \right)}^{2}}\times 1+3\times \dfrac{3}{2}x\times {{\left( 1 \right)}^{2}} \\
 & =\dfrac{27}{8}{{x}^{3}}+1+\dfrac{27}{4}{{x}^{2}}+\dfrac{9}{2}x \\
 & =\dfrac{27{{x}^{3}}+54{{x}^{2}}+36x+8}{8} \\
\end{align}$
Therefore, we get the expanded form of the cube of the given sum of numbers is $\dfrac{27{{x}^{3}}+54{{x}^{2}}+36x+8}{8}$.

Note: In this question, we can also exchange the values of a and b as 1 and $\dfrac{3}{2}x$ and still get the same required answer. This happens as the Commutative Law of addition states that if x and y are two numbers then \[x+y=y+x\] . So, in all the questions of this sort we can do the same thing.