Answer
$(x+1)^2(x^2-x+1)^2$
Work Step by Step
We rewrite the polynomial
$x^6+2x^3+1=y^2+2y+1$
where we made the substitution $y=x^3$. The right side is of the form $a^2+2ab+b^2=(a+b)^2$, which is a complete square. That is, we have
$$
x^6+2x^3+1=y^2+2y+1=(y+1)^2=(x^3+1)^2
.$$
By using the fact that $x^3+1$ is a sum of cubes, we have
$$
x^6+2x^3+1= (x^3+1)^2=(x+1)^2(x^2-x+1)^2
.$$
