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

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Appendix A - Review - A.4 Factoring Polynomials - A.4 Assess Your Understanding - Page A40: 99



Work Step by Step

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 $$ x^6-2x^3+1= (x^3-1)^2=(x-1)^2(x^2+x+1)^2 .$$
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