Let $R=\{\left( a,{{a}^{3}} \right):a\text{ }is\text{ }a\text{ }prime\text{ }number\text{ }less\text{ }than\text{ }10\}$. Find range $\left( R \right)$ .

Hint: To solve such questions first write the prime numbers which are less than $\;10$ . Next for each prime number that is less than $\;10$ find the cube of that prime number. The range will be the set that contains the cube of the prime number.

Complete step-by-step solution:
Given that $R = \{ \left( {a,{a^3}} \right):a{\text{ }}is{\text{ }}a{\text{ }}prime{\text{ }}number{\text{ }}less{\text{ }}than{\text{ }}10\}$
It is said that $a$ is a prime number that is less than $\;10$.
Therefore, the prime numbers that are less than $\;10$ are those written below,
$\;2,3,5,7$
Therefore,
$a = \left\{ {2,3,5,7} \right\}$
The relation is given as $R = \{ a,{a^3}\}$. That is,
If $a = 2$ then ${a^3} = 8$
If $a = 3$ then ${a^3} = 27$
If $a = 5$ then ${a^3} = 125$
If $a = 7$ then ${a^3} = 343$
Therefore, the relation can be written as
$R = \left\{ {\left( {2,8} \right),\left( {3,27} \right),\left( {5,125} \right),\left( {7,343} \right)} \right\}$

Hence the range can be written as, Range $= \left\{ {8,27,125,343} \right\}$

Additional Information: A Relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product set $A \times B$. A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. Those numbers that have only two factors that are the one and the number itself are known as prime numbers. To represent the domain and range of a function one can use interval notation or can also use inequalities.

Note: To solve such questions remember that domain is the set of all possible values of a function and range is the set of all possible output of the function. To compute the domain and the range of a given function, first, we have to find all the possible values that satisfy the given function and then find the output of the function based on these values.