Question

How do you factor \[{x^8} - 1\]?

Answer 1

Hint: We write the given equation in such a way that we can open it using the identity \[{a^2} - {b^2} = (a - b)(a + b)\], first opening helps us to bring power of x from 8 to 4, then again apply same method and power of x from 4 to 2. At the end we obtain power of x from 2 to 1 as well.
* Factor of an equation means that the equation is exactly divisible by that factor. If there is one factor of an equation, then there can be other factors as well which when multiplied to each other give us the original equation.

Complete step-by-step answer:
We have to find factors of \[{x^8} - 1\]
We have to break the term \[{x^8} - 1\] in the product of its factors such that when multiplied, they give us the value \[{x^8} - 1\].
Let us write power 8 as product of 4 and 2.
\[ \Rightarrow {x^8} - 1 = {\left( {{x^4}} \right)^2} - {\left( 1 \right)^2}\]
Now we can apply the identity \[{a^2} - {b^2} = (a - b)(a + b)\]in right hand side of the equation
\[ \Rightarrow {x^8} - 1 = \left( {{x^4} - 1} \right)\left( {{x^4} + 1} \right)\]
Now again we can write the power 4 as product of 2 and 2
\[ \Rightarrow {x^8} - 1 = \left( {{{\left( {{x^2}} \right)}^2} - {{\left( 1 \right)}^2}} \right)\left( {{x^4} + 1} \right)\]
Now we can apply the identity \[{a^2} - {b^2} = (a - b)(a + b)\]in right hand side of the equation
\[ \Rightarrow {x^8} - 1 = \left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)\]
Now we can apply the identity \[{a^2} - {b^2} = (a - b)(a + b)\]in right hand side of the equation
\[ \Rightarrow {x^8} - 1 = \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)\]

\[\therefore \]The factors of \[{x^8} - 1\]are \[\left( {x - 1} \right);\left( {x + 1} \right);\left( {{x^2} + 1} \right);\left( {{x^4} + 1} \right)\].

Note:
Many students make mistake of breaking the factors in addition as well with the same identity as they write negative sign in between and one negative sign with 1, but keep in mind square of -1 will always give 1, so there is no use of these manipulations, these values in addition are directly factors of the given equation as they divide the given equation completely.