Question

How do you factor $({x^6}) + 1$?

Answer 1

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.