Factorize the algebraic expression: \[{x^2} - 1 - 2a - {a^2}\]

Hint: To factorize the given expression we should first take out minus one common from the last three terms. Then further solve the bracket term by using the identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].After this our expression becomes in the form of \[{a^2} - {b^2}\]. Therefore, we will replace it with \[\left( {a + b} \right)\left( {a - b} \right)\]. On further solving we get the desired solution.

Complete step-by-step solution:
Expression given in the question is \[{x^2} - 1 - 2a - {a^2}\].We have to factorize this expression.
\[ \Rightarrow {x^2} - 1 - 2a - {a^2}\]
Our first step is to take out \[ - 1\] common from the last three terms. By doing this the above expression becomes
\[ \Rightarrow {x^2} - \left( {1 + 2a + {a^2}} \right)\]
We know that \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] which holds for all the values of a and b. This identity can be used as a shortcut. In simple words, \[{\left( {a + b} \right)^2}\] can be replaced by \[{a^2} + {b^2} + 2ab\] and vice versa in the numericals. Now observe the expression, \[{x^2} - \left( {1 + 2a + {a^2}} \right)\]. We can write the bracket term in the form of \[{a^2} + {b^2} + 2ab\] as
\[ \Rightarrow {x^2} - \left( {{{\left( 1 \right)}^2} + {a^2} + 2\left( 1 \right)\left( a \right)} \right)\]
And we know that \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] .So replace the bracket by the term in the form of \[{\left( {a + b} \right)^2}\] as here a is equal to one and b is equal to a. Therefore, the above expression becomes
\[ \Rightarrow {x^2} - {\left( {1 + a} \right)^2}\]
There is one more identity, \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] .That is the special product of the binomials a+b and a-b is equal to the difference of squares of the terms a and b. In the expression \[{x^2} - {\left( {1 + a} \right)^2}\], x square is equal to a square and \[{\left( {1 + a} \right)^2}\] is equal to b square. Therefore,
 \[ \Rightarrow \left( {x + \left( {1 + a} \right)} \right)\left( {x - \left( {1 + a} \right)} \right)\]
On expanding the above expression we get
\[ \Rightarrow \left( {x + 1 + a} \right)\left( {x - 1 - a} \right)\]
Hence, on doing factorization of \[{x^2} - 1 - 2a - {a^2}\] we get \[\left( {x + 1 + a} \right)\left( {x - 1 - a} \right)\].

Note: Keep in mind that the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] is mainly used to simplify an expression when the expression satisfies the condition: First, the two binomials should be formed by the two variables. Second, the two variables should be connected with opposite signs in the binomials.