How do you simplify $ {\left( {\dfrac{5}{8}} \right)^2} $ ?

Hint: Here we are given one fraction and the whole square to the fraction, where fraction is the term which is expressed as the number where the ratio of two numbers in the numerator is to the denominator. Also, here we will use the identity and apply square to both the terms in the numerator and the denominator.

Complete step-by-step answer:
Take the given expression –
 $ {\left( {\dfrac{5}{8}} \right)^2} $
By using the identity, when there is a square applied to the whole bracket we can apply to both the terms in the given fraction.
 $ \Rightarrow {\left( {\dfrac{5}{8}} \right)^2} = \dfrac{{{5^2}}}{{{8^2}}} $
Now, square is the number which is multiplied with itself twice.
 $ \Rightarrow {\left( {\dfrac{5}{8}} \right)^2} = \dfrac{{5 \times 5}}{{8 \times 8}} $
Now simplify the above expression –
 $ \Rightarrow {\left( {\dfrac{5}{8}} \right)^2} = \dfrac{{25}}{{64}} $
This is the required solution.
So, the correct answer is “ $ \dfrac{{25}}{{64}} $ ”.

Note: cube is the product of same number three times such as $ {n^3} = n \times n \times n $ for Example cube of $ 2 $ is $ {2^3} = 2 \times 2 \times 2 $ simplified form of cubed number is $ {2^3} = 2 \times 2 \times 2 = 8 $ . and cube-root is denoted by $ \sqrt[3]{{{n^3}}} = \sqrt {n \times n \times n} = n $ For Example: $ \sqrt[3]{8} = \sqrt[3]{{{2^3}}} = 2 $ Do not be confused in square and square-root similarly cubes and cube-root, know the concepts properly and apply accordingly.