I just want to understand how you get the three when you have to cube three binomials together.

my tutor gave me this formula to follow when i come across anything I have to cube. However I'm not sure how she got the 3 there.

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- Jun 1st 2011, 09:20 AMsara213quick question about cubing polynomials.
I just want to understand how you get the three when you have to cube three binomials together.

my tutor gave me this formula to follow when i come across anything I have to cube. However I'm not sure how she got the 3 there. - Jun 1st 2011, 09:26 AMPlato
- Jun 1st 2011, 09:42 AMsara213
- Jun 1st 2011, 09:47 AMPlato
- Jun 1st 2011, 09:50 AMtopsquark
His point is that there are terms (a)(a)(b), (a)(b)(a), and (b)(a)(a) that come up. Since there are three of them, you get a 3 in front of the .

However let's look at this differently. , right? Then you expand and simplify to get . The 2 in front of the ab appears because we have a term ab and ba.

Similarly:

Expand this and simplify. Here you will find, for instance, that you have a and a which gives you .

-Dan - Jun 1st 2011, 09:51 AMe^(i*pi)
The order doesn't matter in multiplication - 5*3 is the same as 3*5. This extends to letters too so and since you have three lots in your expansion it gives

If "proving" it isn't all that important look up Pascal's triangle - that gives you the coefficients necessary. Or, better yet, the binomial theorem - Jun 1st 2011, 10:04 AMsara213
Okay got it...sorry plato, I did not understand what you are trying to say in the first post.