Solve the factor ${x^3} + \dfrac{1}{8}$?

Hint: In the given question, we have been given a polynomial. The degree of the polynomial is three. We have to factorize the given polynomial. The first part of the polynomial is a cube of the variable. While the second part of the polynomial is a number. And, this number is also a cube. The two parts of the polynomial are added by a basic arithmetic operator the plus “+” sign. So, we can easily factorize the given polynomial by applying the formula of the sum of two cubes.

Formula Used: -
In the given question, we are going to apply the formula of the sum of two cubes.
Let there be two integral numbers; say the numbers are $a$ and $b$, the sum of their cubes can be written as:
${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$

Complete step by step answer:
In the given question, we are going to apply the formula of the sum of two cubes.
Let there be two integral numbers; say the numbers are $a$ and $b$.
Now, there is a standard formula for the factorization of the sum of two cubes, which is:
${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$
Substitute the values $a = x$ and $b = \dfrac{1}{2}$, we get
$ \Rightarrow {x^3} + {\left( {\dfrac{1}{2}} \right)^3} = \left( {x + \dfrac{1}{2}} \right)\left( {{x^2} - x \times \dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right)$
Simplify the terms,
$ \Rightarrow {x^3} + \dfrac{1}{8} = \left( {x + \dfrac{1}{2}} \right)\left( {{x^2} - \dfrac{x}{2} + \dfrac{1}{4}} \right)$
Hence, the factor of ${x^3} + \dfrac{1}{8}$ is $\left( {x + \dfrac{1}{2}} \right)\left( {{x^2} - \dfrac{x}{2} + \dfrac{1}{4}} \right)$.

Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. In the given question, we were given a polynomial. We first found what type of polynomial it is. Then we wrote down the formula which can be used to solve or factorize the given polynomial. Then we just put in the values from the question and just solve it like normal.