Question

How do you factor ${{x}^{6}}-1$?

Answer 1

Hint: Factors are the numbers that are multiplied to get another number. For example , \[2\] and $3$ are the factors of \[6\] because\[6=2\times 3\]. Factorization is the way of finding out the factors of the numbers. Factorization can be done in more than one way.

Complete step by step solution:
The given polynomial is ${{x}^{6}}-1$.
It can also be written as
\[{{x}^{6}}-1={{({{x}^{3}})}^{2}}-{{1}^{2}}\]
There are several factorization formulas required to factor a difference or a sum of binomials. The first equation corresponds to a difference of squares and is given by
\[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b).\]
Using this identify we can write the polynomial ${{({{x}^{3}})}^{2}}-{{1}^{2}}$ as
\[{{({{x}^{3}})}^{2}}-{{1}^{2}}=({{x}^{3}}-1)({{x}^{3}}+1)\]
Additionally, a difference or a sum of two cubes would be factored using the formulas given below the following identities
\[\begin{align}
  & {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}}), \\
 & {{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}-ab+{{b}^{2}}) \\
\end{align}\]
Using these identities in the above equation , we can write
\[{{x}^{3}}-1={{x}^{3}}-{{1}^{3}}=(x-1)({{x}^{2}}+x\times 1+{{1}^{2}})=(x-1)({{x}^{2}}+x+1)\]
and \[{{x}^{3}}+1={{x}^{3}}+{{1}^{3}}=(x+1)({{x}^{2}}-x\times 1+{{1}^{2}})=(x+1)({{x}^{2}}-x+1)\]
Putting the value of \[{{x}^{3}}-1\] and ${{x}^{3}}+1$in the above equation , we get
\[\begin{align}
  & {{x}^{6}}-1={{({{x}^{3}})}^{2}}-{{1}^{2}}=({{x}^{3}}-1)({{x}^{3}}+1) \\
 & =(x-1)({{x}^{2}}+x+1)(x+1)({{x}^{2}}-x+1) \\
 & =(x-1)(x+1)({{x}^{2}}+x+1)({{x}^{2}}-x+1) \\
\end{align}\]

Therefore the factors of \[{{x}^{6}}-1\] are \[(x-1),(x+1),({{x}^{2}}+x+1)and({{x}^{2}}-x+1)\].

Note: There are divisibility rules to find the factors of a number. For example, if a number is divisible by two, then it is divisible by \[4\], i.e. \[4\] is a factor, and if a number is divisible by $2$ and by \[3\], then it is also divisible by $6$. The quickest way to find the factors of a number is to divide it by the smallest prime number (bigger than \[1\]) that goes into it evenly with no remainder. Continue this process with each number you get, until you reach \[1\].