How do you factor the expression $ {x^8} - 1 $ ?

Hint: As we know that factorising is the reverse of expanding brackets, it is an important way of solving equations. The first step of factorising an expression is to take out any common factors which the terms have. So if we were asked to factor the expression $ {x^2} + x $ , since $ x $ goes into both terms, we would write $ x(x + 1) $ . Here we will use identities which will help us to factorise an algebraic expression easily i.e. $ ({a^2} - {b^2}) = {a^2} - 2ab + {b^2} $ and $ {a^2} - {b^2} = (a + b)(a - b) $ .

Complete step by step answer:
Here we will use some identities to help like the difference of square identity: $ {a^2} - {b^2} = (a + b)(a - b) $ .
We can further write it as $ {x^8} - 1 = {({x^4})^2} - {1^2} $ , based on the above identity we get $ {x^8} - 1 = ({x^4} + 1)({x^4} - 1) $ and further using same identity we get,
 $ {x^8} - 1 = ({x^4} + 1)\{ {({x^2})^2} - {1^2}\} $ $ \Rightarrow $ $ {x^8} - 1 = ({x^4} + 1)({x^2} + 1)({x^2} - 1) $ . So we get
 $ {x^8} - 1 = ({x^4} + 1)({x^2} + 1)({x^2} - 1) $ . Again, By expanding the terms using the formula we get ,
 $ {x^8} - 1 = \{ {({x^2})^2} + {1^2} + 2{x^2} - 2{x^2}\} ({x^2} - 1)(x + 1)(x - 1) $ ,
 On further simplifying the above expression by applying the sum identity we get that,
 $ {x^8} - 1 = [\{ {({x^2})^2} + {1^2} + 2{x^2}\} - 2{x^2}]({x^2} + 1)(x + 1)(x - 1) $ .
Applying the formula in the equation we get that , $ {x^8} - 1 = ({x^2} + 1 + \sqrt 2 x)({x^2} + 1 - \sqrt 2 x)({x^2} + 1)(x + 1)(x - 1) $ .
Hence the answer of $ {x^8} - 1 = ({x^2} + 1 + \sqrt 2 x)({x^2} + 1 - \sqrt 2 x)({x^2} + 1)(x + 1)(x - 1) $ .

Note:
 We should keep in mind while solving this expression that we use correct identities to factorize the given algebraic expressions and keep checking the negative and positive signs otherwise it will give the wrong answer. Also we should know that $ ({a^2} + {b^2}) = {a^2} + 2ab + {b^2} $ is the identity that is also used in the above solution. These are some of the standard algebraic identities. This is as far we can go with real coefficients as the remaining quadratic factors all have complex zeroes.