(a) The expression $x^5 + y^5$ can be written as the product of $x+y$ and another factor. Find that other factor.
$x^5 - y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$.
We can use this established result for $x^5 + y^5$
$x^5 + y^5 = x^5 - (-y)^5 = [(x-(-y)][x^4+x^3(-y)+x^2(-y)^2+x(-y)^3+(-y)^4]=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)$.
(b) The expression $x^7 + y^7$ can be written as the product $x+y$ and another factor. Find that other factor.
$x^7 - y^7 = (x-y)(x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+xy^5+y^6)$.
Once again using the established result we find, $x^7 + y^7 = (x+y)(x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)$.
(c) Write $x^{2n+1} + x^{2n+1}$ as the product of two factors.
\begin{align*} {x^{2n+1} + x^{2n+1}}&=(x+y)(x^{2n}+x^{2n-1}(-y)+\cdots + x(-y)^{2n-1}+(-y)^{2n})\\ &\,\,{=(x+y)\sum_{k=0}^{2n} x^{2n-k}(-y)^k} \end{align*}
(d) Why does the factorization in the previous part fail when the powers of $x$ and $y$ are even? In other words, why can we not factor $x^4+y^4$ or $x^6 + y^6$ using the patterns we found in the first three parts?
$x^4 - y^4 = (x-y)(x^3+x^2y+xy^2+y^3)$
$x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$
When the powers are even, the factorization takes on a different form where the numbers of terms are even instead of odd as seen the earlier parts. The highest degree is also even so we end up having $x^n-y^n$ when we try to factor in the same way.
I'm still trying to figure out the factorization of $x^4+y^4$ and $x^6 + y^6$.