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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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    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.2011-02-04

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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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    Me too! I think I may star them all!2012-07-16
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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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    @Tom Brito: And you may want to refresh yourself on Polynomial Long Division: http://en.wikipedia.org/wiki/Polynomial_long_division2019-05-15
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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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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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    But if you go right to left, all the words are backwards. I imagine this is where the confusion is coming from.2012-01-25
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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

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