How do you simplify \[{\left( {27} \right)^{\dfrac{1}{3}}}\]?

Hint: In the given question, we have been given an expression. First, let us rewrite the expression as \[\sqrt[3]{{27}}\]. The expression involves a natural number inside an integral root bracket. We have to evaluate the value of the expression. To do that, we have to form a collection of as many numbers as is the value of the root. Then, if we have any such collection, we take them out of the root bracket and if there are some numbers which do not fit in it, they remain in the bracket.

Formula Used:
In this question, we are going to use the formula of cube.
Say there is a number \[a\], then its cube can be represented as
\[{a^3}\]

Complete step-by-step answer:
The given expression to be simplified in this question is \[{27^{\dfrac{1}{3}}} = \sqrt[3]{{27}}\], i.e., we have to calculate the cube root of \[27\].
For solving it, we are first going to do the prime factorization of \[27\], combine the factors in the form of triplets of equal numbers, take out the triplets as a single number, leave the numbers which do not combine as triplet in the root, and that is going to give us the answer.
\[\begin{array}{l}3\left| \!{\underline {\,
  {27} \,}} \right. \\3\left| \!{\underline {\,
  9 \,}} \right. \\3\left| \!{\underline {\,
  3 \,}} \right. \\{\rm{ }}\left| \!{\underline {\,
  1 \,}} \right. \end{array}\]
So, \[27 = \times 3 \times 3 \times 3 = {3^3}\]
Hence, \[{27^{\dfrac{1}{3}}} = \sqrt[3]{{27}} = \sqrt[3]{{{{\left( 3 \right)}^3}}} = 3\]
Additional Information:
In the given question, the representation of the cube root was given as \[{27^{\dfrac{1}{3}}}\]. But if it was given as \[\sqrt[3]{{27}}\], then it means the same thing – cube root, and is just a way to represent it differently.

Note: When we are calculating such questions, we find the prime factorization, club the triplets together, take them out as a single number and solve for them. This procedure requires no further action or steps to evaluate the answer. It is a point to remember that a perfect cube always has triplets of factors.