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

2 Answers
May 16, 2015

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!

May 17, 2015

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