1. **State the problem:** Simplify or factor the expression $$y^4 + y^2 - 2ay + 1 - a^2$$. 2. **Recall the formula:** Recognize that $$1 - a^2$$ can be written as $$(1 - a)(1 + a)$$, and look for ways to factor the polynomial by grouping or substitution. 3. **Rewrite the expression:** Group terms to see if it fits a pattern: $$y^4 + y^2 - 2ay + 1 - a^2 = (y^4 + y^2) - 2ay + (1 - a^2)$$ 4. **Try substitution:** Let’s consider the expression as a quadratic in $y^2$ or try to factor as a product of two quadratics: Assume: $$(y^2 + py + q)(y^2 + ry + s) = y^4 + y^2 - 2ay + 1 - a^2$$ 5. **Expand the assumed product:** $$y^4 + (p + r) y^3 + (pr + q + s) y^2 + (ps + rq) y + qs$$ 6. **Match coefficients:** - Coefficient of $y^4$ is 1, matches. - Coefficient of $y^3$ is 0, so $p + r = 0$. - Coefficient of $y^2$ is 1, so $pr + q + s = 1$. - Coefficient of $y$ is $-2a$, so $ps + rq = -2a$. - Constant term is $1 - a^2$, so $qs = 1 - a^2$. 7. **From $p + r = 0$, let $r = -p$**. 8. **Rewrite the $y^2$ coefficient:** $$pr + q + s = p(-p) + q + s = -p^2 + q + s = 1$$ 9. **Rewrite the $y$ coefficient:** $$ps + r q = p s - p q = p(s - q) = -2a$$ 10. **Rewrite the constant term:** $$q s = 1 - a^2$$ 11. **Try $q = 1 - a$ and $s = 1 + a$ (since their product is $1 - a^2$):** 12. **Calculate $s - q = (1 + a) - (1 - a) = 2a$** 13. **From $p(s - q) = -2a$, substitute:** $$p imes 2a = -2a \\ p = -1$$ 14. **From $-p^2 + q + s = 1$, substitute:** $$-(-1)^2 + (1 - a) + (1 + a) = -1 + 1 - a + 1 + a = 1$$ 15. **All conditions are satisfied.** 16. **Therefore, the factorization is:** $$ (y^2 - y + 1 - a)(y^2 + y + 1 + a) $$ **Final answer:** $$y^4 + y^2 - 2ay + 1 - a^2 = (y^2 - y + 1 - a)(y^2 + y + 1 + a)$$