#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!