# Signs in binomial expansions

Edit the title as seems fit.

\begin{align} (a^3+b^3) &= (a+b)(a^2 -ab+b^2) \\ &= (a+b)^3 -3ab(a+b) \end{align}

And so on and so forth. Right now, I only need these expansions in solving quadratic equations. But why do signs vary in the expansions? (asterisk). What controls this? I see that something similar comes in $a^2-b^2 = (a+b)(a-b)$ to allow the intermediate term(s) to cancel but how does this translate to other (higher-order) forms?

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The way you're using words to ask this question isn't translating well. I think you may have a valid question. But, the current choice of words makes it very hard to understand. – 000 May 12 '12 at 21:22

One way of looking at this is to see that you need the intermediate terms to cancel, so taking out a factor of $(a+b)$ you will need alternating signs for the cancellation to work.
That doesn't tell why it was decided (apparently, that's how I was given it) that such terms, at face value, need the expansion with opposite signs. I added an (*) in the \$\$ so see there for what I meant. – Noein May 12 '12 at 21:03
I'm not at all sure what you mean by "it was decided" - do the multiplication. One way works and the signs cancel. The other way doesn't and they don't. Write it out as a multiplication - multiply by $x$ first, and put this in the top line on our paper, and then by $y$ on the next line - making sure you put the terms with the same power of $x$ under/over each other, and you will see the logic. – Mark Bennet May 12 '12 at 21:10
Or do the above with $a$ and $b$ instead of $x$ and $y$ – Mark Bennet May 12 '12 at 21:18
Suppose you have a homogeneous expression in $x$ and $y$ with consecutive terms $Ax^{r+1} y^s+Ax^r y^{s+1}$ and you multiply by $x-y$. Just look at the term in $x^{r+1} y^{s+1}$ - you get a positive contribution by multiplying the second monomial in the first expression by $+x$ and an "equal and opposite" negative contribution by multiplying the first monomial by $-y$ - it works because the coefficients are the same and the signs in the expression $x-y$ are different. I'm doing my best to explain. – Mark Bennet May 17 '12 at 20:39