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August 31st, 2008, 02:13 PM  #1 
Newbie Joined: Aug 2008 Posts: 16 Thanks: 0  The Sum of Cubes
I understand that Code: A^3 + B^3 = (A + B)(A^2  AB + B^2). (A + B)^3 and how it would end up like: (A^3 + 3A^2B + 3AB^2 + B^3). I know (A + B)^3 ends up as (A + B)(A + B)(A + B). But I can't quite understand the (A + B)(A^2  AB + B^2) formula become A^3 + B^3. I also need help in recognizing (A + B)(A^2  AB + B^2) and know that it is the Sum of cubes and not the factor of cubes. Any help would be appreciated. Thanks! 
August 31st, 2008, 02:40 PM  #2 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0  Re: The Sum of Cubes
Hi Do you know how to multiply out brackets with more than two terms in? The identity can be shown directly just by multiplying out : 
August 31st, 2008, 02:48 PM  #3 
Newbie Joined: Aug 2008 Posts: 16 Thanks: 0  Re: The Sum of Cubes
Ah, Now I see the get canceled out. Thanks!

September 1st, 2008, 04:17 PM  #4 
Newbie Joined: Aug 2008 Posts: 16 Thanks: 0  Re: The Sum of Cubes
Is there any way to get without having to remember that specific formula? I am looking for a way to solve the same way I would solve something like by using the FOIL method to get the proper answer and not having to remember that it equals 
September 1st, 2008, 06:32 PM  #5  
Senior Member Joined: Jul 2008 Posts: 895 Thanks: 0  Re: The Sum of Cubes Quote:
As pointed out, multiply the two to find that it is an identity. Try the reverse by long division. Look also at geometric series and fit it into that picture. It's no worse than seeing numbers for the first time an having to memorise processes in arithmetic as a child. That's why it is a step up in your education. You have the familiar extended to the more abstract. You know that (2)(3) = 6, 3 = 6/2, 2 = 6/3 and so on by similar familiarity, and that (2) and (3) are factors of 6. But FIRST you have things to memorise. What will really embed it is practice: a^3  b^3 = x^3 + y^3 = p^(3/2) + q^(3/2) = and so on. Keep going until certain of the pattern.  

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