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Is there a way of go from $a^3+b^3$ to $(a+b)(a^2-ab+b^2)$ other than know the property by heart?

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Long division of $a^3+b^3$ by $a+b$? –  Arturo Magidin Feb 3 '11 at 23:12
If you are asking how to arrive from one to another without knowing that in the first place, no it is not possible. Just like everything the more your work with things the more you'll know about them. But now that you know the above identity, see what makes it work see if you can come up with something for $a^2+b^2$ knowing what you know. It is all in trying and trying and..., and trying different things. –  Arjang Feb 4 '11 at 2:07
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5 Answers

up vote 18 down vote accepted

If you want to know if $a^n + b^n$ is divisible by $a+b$ (or by $a-b$, perhaps), you can always try long division, whether explicitly or in your head. I can't figure out a way to do the LaTeX or ASCII art for it here to do it explicitly, so I'll show you how one would do it "in one's head".

For example, for $a^3+b^3$, to divide $a^3+b^3$ by $a+b$, start by writing $a^3+b^3= (a+b)(\cdots)$. Then: we need to multiply $a$ by $a^2$ to get the $a^3$, so we will get $a^3+b^3=(a+b)(a^2+\cdots)$. The $a^2$ makes an unwanted $a^2b$ when multiplied by $b$, so we need ot get rid of it; how? We multiply $a$ by $-ab$. So now we have $$a^3+b^3 = (a+b)(a^2-ab+\cdots).$$ But now you have an unwanted $-ab^2$ you get when you multiply $b$ by $-ab$; to get rid of that $-ab^2$, you have to "create" an $ab^2$. How? We multiply $a$ by $b^2$. So now we have: $$a^3 + b^3 = (a+b)(a^2-ab+b^2+\cdots)$$ and then we notice that it comes out exactly, since we do want the $b^3$ that wee get when we multiply $b^2$ by $b$; so $$a^3 + b^3 = (a+b)(a^2-ab+b^2).$$

If the expression you want is not divisible by what you are trying, you'll run into problems which require a "remainder". For instance, if you tried to do the same thing with $a^4+b^4$, you would start with $a^4+b^4 = (a+b)(a^3+\cdots)$; then to get rid of the extra $a^3b$, we subtract $a^2b$: $a^4+b^4 = (a+b)(a^3 - a^2b+\cdots)$. Now, to get rid of the unwanted $-a^2b^2$, we add $ab^2$, to get $a^4+b^4 = (a+b)(a^3-a^2b+ab^2+\cdots)$. Now, to get rid of the unwanted $ab^3$, we subtract $b^3$, to get $$a^4+b^4 = (a+b)(a^3-a^2b+ab^2 - b^3+\cdots).$$ At this point we notice that we get an unwanted $-b^4$ (unwanted because we want $+b^4$, not $-b^4$). There is no way to get rid of it with the $a$, so this will be a "remainder". We need to cancel it out (by adding $b^4$) and then add what is still missing (another $b^4$), so $$a^4 + b^4 = (a+b)(a^3-a^2b+ab^2 -b^3) + 2b^4.$$ (Which, by the way, tells you that $a^4-b^4 = (a+b)(a^3-a^2b+ab^2-b^3)$, by moving the $2b^4$ to the left hand side).

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I love how simple, yet very helpful your explanations are. –  daniel.jackson Feb 4 '11 at 9:44
Me too! I think I may star them all! –  Ziggy Jul 16 '12 at 19:22
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Keeping $a$ fixed and treating $a^3 + b^3$ as a polynomial in $b$, you should immediately notice that $-a$ will be a root of that polynomial. This tells you that you can divide it by $a + b$. Then you apply long division as mentioned in one of the comments to get the other factor.

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sorry, but how's the procedure to divide $a^3+b^3$ for $a+b$? –  Tom Brito Feb 3 '11 at 23:32
Look at the answer of Arturo Magidin above. :) –  Lagerbaer Feb 3 '11 at 23:55
@Tom Brito: And you may want to refresh yourself on Polynomial Long Division: en.wikipedia.org/wiki/Polynomial_long_division –  Arturo Magidin Feb 4 '11 at 0:10
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It's a homogenization of the cyclotomic factorization $\rm\ x^3 + 1 = (x+1)\ (x^2 - x + 1)\:.\ $ Recall that the homogenization of a degree $\rm\:n\:$ polynomial $\rm\ f(x)\ $ is the polynomial $\rm\ y^n\ f(x/y)\:.\ $ This maps $\rm\ x^{k}\ \to\ x^{k}\ y^{n-k}\ $ so the result is a homogeneous polynomial of degree $\rm\:n\:.\ $ While this cyclotomic factorization is rather trivial, other cyclotomic homogenizations can be far less trivial. For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:

$$\begin{array}{rl} x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\ \frac{x^6 + 3^2}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\ \frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\ \frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\ \end{array}$$

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lol-cakes...... –  Adam Jan 25 '12 at 8:43
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So I guess the answer to this question is to expand the right-hand side of the equation, although it might not be too clear on how to do this.

It might be clearer if you make the substitution $u=a+b$. Then the right hand side is:

\begin{align*} (a+b)(a^2-ab+b^2) & = u(a^2-ab+b^2) \\ & = ua^2-uab+ub^2 \\ & = (a+b)a^2-(a+b)ab+(a+b)b^2 \\ & = a^3+ba^2-a^2b-ab^2+ab^2+b^3 \\ & = a^3+b^3 \end{align*} using the distributive law (and commutivity).

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you demonstrated the reverse of what I wanted. –  Tom Brito Feb 3 '11 at 23:29
@Tom: The title of the question is different from what you ask in the contents. The title asks for how to know that they are equal. The contents take another direction and ask how to go from the LHS to the RHS. Douglas was most likely answering the title question, which is very reasonable. –  Mitch Feb 4 '11 at 3:04
Why does it matter which direction you go in? y=x is the same as x=y. Alternatively, you can also read it from bottom to top. –  Douglas S. Stones Feb 4 '11 at 3:25
The two directions are pretty distinct here. The one not intended simply confirms the equality by doing the straightforward multiplication and simplification that you showed. The intended direction is really to factor $a^3+b^3$, which is non-trivial. Confirming an identity is much easier than trying to discover a 'simpler' form like the RHS. –  Mitch Feb 6 '11 at 17:06
But if you go right to left, all the words are backwards. I imagine this is where the confusion is coming from. –  Adam Jan 25 '12 at 8:44
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There is a 'trick' for odd powered exponents as follows;

recall the identity

$ (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}) = x^n - y^n $ (*)

Now, there is a real number, a, such that y = (-a) [really this real number is -y but it is not important for the purposes of searching for a possible factorization].

For n even:

$y^n = (-a)^n = a^n$

but for odd n:

$y^n = (-a)^n = -a^n$

If we substitute -a for y in (*) we obtain (we assume n is odd)

$ x^n - y^n = x^n - (-a^n) = x^n + a^n = (x+a)(x^{n-1}-x^{n-2}a+...-xa^{n-2}+a^{n-1}) $

where we alternate + and - within the last bracket due to the parity of n-i.

When n = 3

$(x-y)(x^2+xy+y^2) = x^3-y^3$

which means

$x^3+y^3 = (x+y)(x^2-xy+y^2)$

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