## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$(x-1)^2(x^2+x+1)^2$
We can rewrite the polynomial $x^6-2x^3+1=y^2-2y+1$ where we substitute $y=x^3$. 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 difference of cubes, we have $$x^6-2x^3+1= (x^3-1)^2=(x-1)^2(x^2+x+1)^2 .$$