Question

How do you factor `x^6-2x^3+1`?

Answer 1

`x^6-2x^3+1 = (x^3)^2-2(x^3)+1` is of the form `y^2-2y+1` where `y = x^3`.

This quadratic formula in `y` factors as follows:

`y^2-2y+1 = (y-1)(y-1) = (y - 1)^2`

So `x^6-2x^3+1 = (x^3 - 1)^2`

`x^3 - 1 = (x - 1)(x^2 + x + 1)`

So `x^6-2x^3+1 = (x - 1)(x^2 + x + 1)(x - 1)(x^2 + x + 1)`

`= (x - 1)^2(x^2 + x + 1)^2`.

`x^2+x+1` has no linear factors with real coefficients. To check this notice that it is of the form `ax^2 + bx + c`, which has discriminant:

`Delta = b^2 - 4ac = 1^2 - 4*1*1 = 1 - 4 = -3`

Being negative, the equation `x^2+x+1 = 0` has no real roots.

One way of checking the answer is to substitute a value for `x` that is not a root into both sides and see if we get the same result:

Try `x=2`:

`x^6-2x^3+1 = 2^6-2x^3+1`

`= 64-(2xx8)+1 = 64-16+1 = 49`

Compare:

`(x - 1)^2(x^2 + x + 1)^2 = (2-1)^2(2^2+2+1)^2`

`1^2*7^2=49`

Well that worked!

Answer 2

`x^6 - 2x^3 + 1` is fairly easy to factor, because it is a perfect square. How do I know this? It's a trinomial in the form `a^2 + 2ab + b^2`, and all trinomials in that form are perfect squares.

This trinomial is the perfect square of `(x^3 - 1)`. To check my work, I'll work backwards:

`(x^3 - 1)(x^3 - 1)`

`=x^6 - x^3 - x^3 + 1`

`=x^6 - 2x^3 + 1`

So, this trinomial has factors of `1`, `x^3 - 1`, and `x^6 - 2x^3 + 1`.

However, as it has been pointed out to me, `(x^3 - 1)` also has factors. Since it is a binomial of the form `a^3 - b^3`, it can also be written as `(a - b)(a^2 + ab + b^2)`.

So, `(x^3 - 1)` factors into `(x - 1)` and `(x^2 + x + 1)`, which both are prime.

The factors of `x^6 - 2x^3 + 1` are:
`1`
`x-1`
`x^2 + x + 1`
`x^3 - 1`
`x^6 - 2x^3 + 1`

More specifically, the PRIME factorization of `x^6 - 2x^3 + 1` is:
`(x - 1)^2(x^2 + x + 1)^2`