How do you simplify ${{36}^{\dfrac{3}{2}}}$ $?$

Hint: To solve this question we need to know the concept of exponents and powers. . In this question we will prime factorise the number to find the $\dfrac{{{3}^{th}}}{2}$ of the number. We will use the Prime Factorization method to get the factors of the number, this helps in finding the factors of the number easily. We can recheck the solution by solving the answer having its power as $\dfrac{2}{3}$ which we get after solving.

Complete step by step answer:
The question asks us to simplify or evaluate ${{36}^{\dfrac{3}{2}}}$ . This is a question of exponent. We need to convert the number $36$ into all the prime factors. To find the $\dfrac{{{3}^{th}}}{2}$ of the number we write the number 36 as the product of all the prime factors associated with it.
We can find the value of factors by prime factorisation of the given number. So it would be written as,
\[36=2\times 2\times 3\times 3\]
Now, on substituting $36$ with the product of its prime factors, we get:
\[\Rightarrow {{\left( 2\times 2\times 3\times 3 \right)}^{\dfrac{3}{2}}}\]
The above function could be written as
\[\Rightarrow {{\left[ {{\left( 2\times 2\times 3\times 3 \right)}^{\dfrac{1}{2}}} \right]}^{3}}\]
We can evaluate the number in two steps, firstly will be finding the square root and then we will find the cube of the result found.
For finding the square root if we have a number $b=a\times a$, then it is the square root of $b$, $\sqrt{b}=\sqrt{a\times a}$ , which is equal to $a$ . So applying the same, on the given value:
$\Rightarrow \sqrt{2\times 2\times 3\times 3}$
$\Rightarrow 2\times 3$
$\Rightarrow 6$
Now we will cube the result found, on doing this we get:
$\Rightarrow {{6}^{3}}$
$\Rightarrow 216$
$\therefore $ On simplification of ${{36}^{\dfrac{3}{2}}}$ we get $216$.

Note: We can check whether the answer is correct or not. To check this we will find ${{\left( 216 \right)}^{\dfrac{2}{3}}}$ and check if the answer matches the question or not. Let us square the value ${{216}^{\dfrac{2}{3}}}$, the result we get after squaring is
\[\Rightarrow {{216}^{\dfrac{2}{3}}}\]
On expanding we get,
\[\Rightarrow {{\left( {{216}^{\dfrac{1}{3}}} \right)}^{2}}\]
$\Rightarrow {{\left[ {{\left( 6\times 6\times 6 \right)}^{\dfrac{1}{3}}} \right]}^{2}}$
$\Rightarrow {{6}^{2}}$
$\Rightarrow 36$
So the value which we get is the same as the question. So our solving is correct for the question.