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I'm reading an old (1895) textbook on algebra (doing a bit of review), and practicing factoring polynomials. The author started with polynomials where all terms share a common factor, like $4a^2 + 4a = 4a(a+1)$; then the difference of two squares, like $x^2-y^2=(x+y)(x-y)$. Next, he covered the sum of two cubes, which I'm less familiar with, but was able to follow: eg. $x^3+y^3=(x+y)(x^2-xy+y^2)$.

The first practice problem he presents, however, is to factor $x^5+y^5$. I can't figure out how this relates to factoring the sum of two cubes, or how to solve it using the methods he's presented so far, or if it's just a typo. I found an answer online, though that didn't make the purpose of the exercise any clearer.

It also exposed my lack of proficiency with polynomial long division, as I tried to divide $x+y$ into $x^5+y^5$ and got stuck when the terms no longer contained $x$'s (my partial answer was $x^4-x^3+x^2y-xy^2+y^3+?$). Update: this part, at least, has been sorted out. The division should yield: $x^4-x^3y+x^2y^2-xy^3+y^4$.

Another practice problem also contains a fifth-power, $8a^5b^3c^6+m^6$, and the best I've come up with is, $$8a^5b^3c^6+m^6 = (2a^{5/3}bc^2)^3+(m^2)^3 \\ = (2a^{5/3}bc^2 + m^2)(4a^{10/3}b^2c^4-2a^{5/3}b^2c^4-2a^{5/3}bc^2m^2+m^4)$$

As far as I can tell, the equations are valid, but I'm not confident that's what the author was going for (and the book doesn't offer any solutions). Any help is appreciated.

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For $x^5+y^5$ at least, it should be $(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)$ - I assume you just made a small error somewhere – mardat Mar 7 '13 at 22:51
@mardat Right you are! I made an early error that threw things off. Thanks for pointing that out. At least I can still divide :) – ivan Mar 7 '13 at 22:57
up vote 4 down vote accepted

If you are doing Exercise 34 in the Project Gutenberg edition, the answer is that these two problems are typos. You can read the original at Google Books e.g. here: the problems should be

$1$. Factor $x^3+y^3$ (was incorrectly printed as $x^5+y^5$ in the Gutenberg version)

$5$. Factor $8a^3b^3c^6+m^6$ (was incorrectly printed as $8a^5b^3c^6 + m^6$ in the Gutenberg version)

There are solutions to the exercises in the back of the book.

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That's a relief! I'll check out the original. Thanks :) – ivan Mar 7 '13 at 23:20
Whatever the case, factoring $x^5+y^5$ is a nice exercise. @ivan Hint: $(x^3+y^3)(x^2+y^2)$ for easy going. – Sawarnik Feb 19 '14 at 12:44

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