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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    $\begingroup$ 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 $\endgroup$ – mardat Mar 7 '13 at 22:51
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    $\begingroup$ @mardat Right you are! I made an early error that threw things off. Thanks for pointing that out. At least I can still divide :) $\endgroup$ – ivan Mar 7 '13 at 22:57

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.

  • $\begingroup$ That's a relief! I'll check out the original. Thanks :) $\endgroup$ – ivan Mar 7 '13 at 23:20
  • $\begingroup$ 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. $\endgroup$ – Sawarnik Feb 19 '14 at 12:44

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