How do you factor \[({x^6}) + 1\]?

Hint:We can solve this using algebraic identities. We can write \[{x^6} = {\left( {{x^3}} \right)^2}\], because we know that \[{\left( {{x^a}} \right)^b} = {x^{ab}}\]. After that we apply \[{a^2} - {b^2} = (a - b)(a + b)\]. To simplify further we need \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\] and \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\]. Here 1 can be written as \[{1^2}\].

Complete step by step solution:
Given, \[({x^6}) + 1\].

As we know we can write \[{x^6} = {\left( {{x^3}} \right)^2}\] and \[1 = {1^2}\].
Then above becomes,
\[ \Rightarrow ({x^6}) + 1 = {\left( {{x^3}} \right)^2} + {1^2}\]

We apply the algebraic formula \[{a^2} - {b^2} = (a - b)(a + b)\]. Here \[a = {x^3}\] and \[b = 1\]then
we have,
\[ = ({x^3} - 1)({x^3} + 1) - - - - (1)\]

Now we have two terms in the form \[{a^3} - {b^3}\] and \[{a^3} + {b^3}\]. ( We can write 1 as
\[{1^3}\] because one to the power of any number is 1, except 0 because \[{1^0} = 0\])

Now we separately simplify each term.
Now take \[({x^3} - 1) = ({x^3} - {1^3})\]

Applying the algebraic identity, \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\]. Here \[a = x\] and \[b =
1\]then we have,
\[ \Rightarrow ({x^3} - 1) = (x - 1)({x^2} + x + 1)\]

Similarly we take \[({x^3} + 1) = ({x^3} + {1^3})\]

Applying the algebraic identity, \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\]. Here \[a = x\] and \[b
= 1\]then we have,
\[ \Rightarrow ({x^3} + 1) = (x + 1)({x^2} - x + 1)\]

Substituting these values in the equation (1) we have,
\[ \Rightarrow ({x^6}) + 1 = (x - 1)({x^2} + x + 1)(x + 1)({x^2} - x + 1)\]

Hence the factors of \[({x^6}) + 1\] are \[(x - 1)({x^2} + x + 1)(x + 1)({x^2} - x + 1)\]

Note: We know that the highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. The roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. Here the degree of the given polynomial is 6. Hence it will have 6 factors or 6 roots. As we can see in the result we have two quadratic equations \[({x^2} + x + 1)\] and \[({x^2} - x + 1)\]. We cannot solve this using simple factorization. Although we can solve this using quadratic formulas, we will get complex roots. In either case the obtained result is correct.