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!