Monday, July 22, 2019

algebra precalculus - How to know that a3+b3=(a+b)(a2ab+b2)



Is there a way of go from a3+b3 to (a+b)(a2ab+b2) other than know the property by heart?


Answer



If you want to know if an+bn is divisible by a+b (or by ab, 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 a3+b3, to divide a3+b3 by a+b, start by writing a3+b3=(a+b)(). Then: we need to multiply a by a2 to get the a3, so we will get a3+b3=(a+b)(a2+). The a2 makes an unwanted a2b when multiplied by b, so we need ot get rid of it; how? We multiply a by ab. So now we have
a3+b3=(a+b)(a2ab+).
But now you have an unwanted ab2 you get when you multiply b by ab; to get rid of that ab2, you have to "create" an ab2. How? We multiply a by b2. So now we have:

a3+b3=(a+b)(a2ab+b2+)
and then we notice that it comes out exactly, since we do want the b3 that wee get when we multiply b2 by b; so
a3+b3=(a+b)(a2ab+b2).



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 a4+b4, you would start with a4+b4=(a+b)(a3+); then to get rid of the extra a3b, we subtract a2b: a4+b4=(a+b)(a3a2b+). Now, to get rid of the unwanted a2b2, we add ab2, to get a4+b4=(a+b)(a3a2b+ab2+). Now, to get rid of the unwanted ab3, we subtract b3, to get
a4+b4=(a+b)(a3a2b+ab2b3+).
At this point we notice that we get an unwanted b4 (unwanted because we want +b4, not b4). 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 b4) and then add what is still missing (another b4), so
a4+b4=(a+b)(a3a2b+ab2b3)+2b4.
(Which, by the way, tells you that a4b4=(a+b)(a3a2b+ab2b3), by moving the 2b4 to the left hand side).


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