(2a-3b)cubed (i dunno how to make a little 3) I have nooo clue how to even start with this O_O
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Originally Posted by kais (2a-3b)cubed (i dunno how to make a little 3) I have nooo clue how to even start with this O_O You should know several ways to do this, one is to multiply out: another is to expand using Pascalls triangle/binomial theorem: CB
Originally Posted by kais (2a-3b)cubed (i dunno how to make a little 3) I have nooo clue how to even start with this O_O use identity in this case put x= 2a and y= 3b
(a-b)^3=a^3-3ab^2-3ba^2-b^3
Last edited by Amer; October 30th 2009 at 09:41 AM.
Hmm, I'm coming up with 4a squared - 12 ab + 9b still not sure if I'm getting it right though?
Originally Posted by kais Hmm, I'm coming up with 4a squared - 12 ab + 9b still not sure if I'm getting it right though? The final result will contain terms of order 3, so there will be terms containing the cube of and of , as well as terms containing and CB
Originally Posted by kais I'm coming up with 4a squared - 12 ab + 9b still not sure if I'm getting it right though? Post a squared this way : a^2 ; a cubed : a^3 4a^2 - 12ab + 9b^2 is what you get from (2a - b) times (2a - b); now you need to multiply (4a^2 - 12ab + 9b^2) times (2a - b); then you're finished.
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