How do you factor $8{{a}^{3}}+1$ ?

Hint: In this question, we have to find the factors of an algebraic expression. So, we will use the algebraic identity and the basic mathematical rule to get the required solution. As we know, 8 is a cube of 2, thus we will rewrite the given expression. After that, we will apply the algebraic identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ in the new algebraic expression. In the last, we will open the brackets of the algebraic expression, to get the required solution to the problem.

Complete step by step solution:
According to the question, we have to find the factors of an algebraic expression.
So, we will apply the algebraic identity and the basic mathematical rules to get the required result.
The expression given to us is $8{{a}^{3}}+1$ ---------------- (1)
First, we know that 8 is a cube of 2 and 1 is the cube of 1, therefore we will rewrite equation (1), we get
$\Rightarrow {{\left( 2a \right)}^{3}}+{{\left( 1 \right)}^{3}}$
Now, we will apply the algebraic identity ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$ in the above expression, Thus, is on comparing above expression and ${{x}^{3}}+{{y}^{3}}$ , we get $x=2a,$ and $y=1$ , thus on substituting the value of x and y in the formula, we get
$\Rightarrow \left( 2a+1 \right)\left( {{\left( 2a \right)}^{2}}-\left( 2a \right)\left( 1 \right)+{{\left( 1 \right)}^{2}} \right)$
On opening the brackets of the above expression, we get
$\Rightarrow \left( 2a+1 \right)\left( 4{{a}^{2}}-2a+1 \right)$ which is the required answer.
Therefore, for the algebraic expression $8{{a}^{3}}+1$ , its factors are $\left( 2a+1 \right)\left( 4{{a}^{2}}-2a+1 \right)$.

Note: While solving this problem, do mention the formula you are using to avoid confusion and mathematical error. For checking your answer, apply the distributive property in the solution, and thus you will get the expression given in the question, which implies you get the accurate answer.