Factorize: \[{{a}^{3}}-{{b}^{3}}+1+3ab\] ?

Hint: In order to find a solution to this problem, we will expand the expression and simplify the expression into its simple terms. We will use factorization property to factorize our expression and use the relationship identity ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$ and equate the terms from our expression to find the solution.

Complete step by step solution:
The factorization is a process of representing the given expression in the form of a product of two factors. These factors may be either of the variables or numbers or algebraic expressions themselves.
We have our polynomial expression as:
\[{{a}^{3}}-{{b}^{3}}+1+3ab\]
Since we have to factorize it, we will have to use the relationship identity ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$ and compare it with our expression.
Therefore, we have:
\[{{a}^{3}}-{{b}^{3}}+1+3ab\]
Rewriting our expression in terms of our identity, therefore, we have:
\[{{a}^{3}}+\left( -{{b}^{3}} \right)+{{1}^{3}}+3ab\]
Therefore, on comparing with our identity expression, we get:
$x=a$
$y=-b$
$z=1$
Since, we get our terms; therefore, we will the solution as per our identity states:
Therefore, we get our solution as:
$\Rightarrow \left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}+1-a\left( -b \right)-\left( -b \right)\left( 1 \right)-a\left( 1 \right) \right)$
Now, on simplifying our expression, we get:
$\Rightarrow \left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}+1+ab+b-a \right)$
We can write our solution as:
\[{{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}+1+ab+b-a \right)\]

Note: To check if our solution is correct or not, we can expand our final solution expression and see if it matches our problem expression; if it matches then our solution is correct.
We have our solution as:
$\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}+1+ab+b-a \right)$
On expanding our expression, we get:
$\Rightarrow a\left( {{a}^{2}}+{{b}^{2}}+1+ab+b-a \right)-b\left( {{a}^{2}}+{{b}^{2}}+1+ab+b-a \right)+1\left( {{a}^{2}}+{{b}^{2}}+1+ab+b-a \right)$
On simplifying our expression, we get:
\[ \Rightarrow {{a}^{3}}+a{{b}^{2}}+a+{{a}^{2}}b+ab-{{a}^{2}}-{{a}^{2}}b-{{b}^{3}}-b-a{{b}^{2}}-{{b}^{2}}+ab+{{a}^{2}}+{{b}^{2}}+1+ab+b-a\]
On simplifying the common terms and equating it, we get:
\[\Rightarrow {{a}^{3}}+ab-{{b}^{3}}+ab+1+ab\]
On simplifying, we get:
\[\Rightarrow {{a}^{3}}-{{b}^{3}}+1+3ab\]
That is our problem expression.
Therefore, this proves that our solution is correct.
That is,
\[{{a}^{3}}-{{b}^{3}}+1+3ab=\left( a-b+1 \right)\left( {{a}^{2}}+{{b}^{2}}+1+ab+b-a \right)\] is the solution.