Answer
$(x+1)^2(x^2-x+1)^2$
Work Step by Step
With $x^6=(x^3)^2$, then
$x^6+2x^3+1=(x^3)^2+2x^3+1$
Letting $x^3=u$ yields
$=u^2+2u+1$
Rewrite $2u$ as $u+u$
$=u^2+u+u+1$
Group the terms.
$=(u^2+u)+(u+1)$
Factor out the GCF in each group.
$=u(u+1)+1(u+1)$
Factor out $(u+1)$.
$=(u+1)(u+1)$
Back substitute $u=x^3$.
$=(x^3+1)(x^3+1)$
$=(x^3+1)^2$
With $1^3=$< the expression above is eqiuvalent to:
$=(x^3+1^3)^2$
Use special formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=x$ and $b=1$ to obtain:
$=[(x+1)(x^2-x+1)]^2$
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 $(x+1)^2(x^2-x+1)^2$.
