What is ${{3}^{\dfrac{3}{2}}}$ in radical form?

Hint: We solve the given question using the laws of exponents. Firstly, we use the product rule of exponents to split the given exponent. Then, we substitute the values of the exponent and simplify them to get the required result.

Complete step-by-step solution:
We are given an exponent and need to write it in radical form. We will be using the laws of exponents to solve the given question.
A radical is used to represent the particular root of the number denoted by $\sqrt{{}}$ . The radical expression must contain a radical sign $\sqrt{{}}$ .
Any expression in the radical form must contain a radical sign that does not hold any perfect squares in it.
According to the question,
The exponent given is ${{3}^{\dfrac{3}{2}}}$
From the laws of exponents,
$\Rightarrow {{a}^{\left( m+n \right)}}={{a}^{m}}\times {{a}^{n}}$
The fraction $\dfrac{3}{2}$ can be written as the sum of 1 and $\dfrac{1}{2}$
Writing the above in the form of equation,
$\Rightarrow \dfrac{3}{2}=1+\dfrac{1}{2}$
Applying the law of exponents, we get,
$\Rightarrow {{3}^{\dfrac{3}{2}}}={{3}^{1}}\times {{3}^{\dfrac{1}{2}}}$
From the law of exponents,
$\Rightarrow {{a}^{\dfrac{1}{m}}}=\sqrt[m]{a}$
Applying the same for the exponent ${{3}^{\dfrac{1}{2}}}$ ,
$\Rightarrow {{3}^{\dfrac{1}{2}}}=\sqrt[2]{3}$
The value of $\sqrt[2]{a}$ can also be written as $\sqrt{a}$
Substituting the same, we get,
$\Rightarrow {{3}^{\dfrac{1}{2}}}=\sqrt{3}$
We need to substitute the value of ${{3}^{\dfrac{1}{2}}}$ in the above expression
$\Rightarrow {{3}^{\dfrac{3}{2}}}={{3}^{1}}\times \sqrt{3}$
Any number to the power of one is the number itself.
$\Rightarrow {{a}^{1}}=a$
From the above,
$\Rightarrow {{3}^{1}}=3$
Substituting the same, we get,
$\Rightarrow {{3}^{\dfrac{3}{2}}}=3\times \sqrt{3}$
The value of the exponent ${{3}^{\dfrac{3}{2}}}$ is written as the product of the number 3 and square root of 3.
$\therefore {{3}^{\dfrac{3}{2}}}=3\sqrt{3}$

Note: The result of the above in radical form can be cross checked as follows,
LHS:
$\Rightarrow {{3}^{\dfrac{3}{2}}}$
RHS:
$\Rightarrow 3\times \sqrt{3}$
$\Rightarrow 3\times {{\left( 3 \right)}^{\dfrac{1}{2}}}$
$\Rightarrow {{\left( 3 \right)}^{1}}\times {{\left( 3 \right)}^{\dfrac{1}{2}}}$
From the laws of exponents,
$\Rightarrow {{a}^{m}}\times {{a}^{n}}={{a}^{\left( m+n \right)}}$
Applying the same, we get,
$\Rightarrow {{\left( 3 \right)}^{1+\dfrac{1}{2}}}$
$\Rightarrow {{3}^{\dfrac{3}{2}}}$
$\therefore$ LHS=RHS, the result attained is correct.