Answer
$(x-1)^2(x^2+x+1)^2$
Work Step by Step
Since $x^6=(x^3)^2$, then we have:
$x^6-2x^3+1=(x^3)^2-2x^3+1$
Letting $x^3=u$ yields
$=u^2-2u+1$
The quadratic trinomial above has $a=1$$b=-2$, and $c=1$
Since $a=1$, look for factors of $c=1$ whose sum is $-2$.
These factors are $-1$ and $-1$.
Use the factors found above to rewrite the middle term.
Rewrite $-2u$ as $-u-u$ to obtain:
$u^2-2u+1=u^2-u-u+1$
Group the terms.
$=(u^2-u)+(-u+1)$
Factor out the GCF in each group.
$=u(u-1)-1(u-1)$
Factor out the GCF $(u-1)$.
$=(u-1)(u-1)$
Back substitute $a=x^3$.
$=(x^3-1)(x^3-1)$
$=(x^3-1)^2$
With $1^3=1$< then the expression above is equivalent to:
$=(x^3-1^3)^2$
Use special formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x$ and $b=1$.
$=[(x-1)(x^2+x(1)+1^2)]^2$
$=[(x-1)(x^2+x+1)]^2$
Use law of the exponents $(ab)^2=a^2b^2$.
$=(x-1)^2(x^2+x+1)^2$
Hence, the completely factored form of the given expression is $(x-1)^2(x^2+x+1)^2$.
