My Math Forum The Sum of Cubes

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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). But Why? I don't understand how the - AB comes into play. I am used to: (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 $(A+B)(A^2-AB+B^2)$: $(A+B)(A^2-AB+B^2)=(A+B)A^2-(A+B)AB+(A+B)B^2\\ \hspace{172mm}=A^3+A^2B-A^2B-AB^2+AB^2+B^3\\ \hspace{172mm}=A^3+B^3.$
 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 $A^2B & AB^2$ 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 $(A + B)(A^2 - AB + B^2)$ without having to remember that specific formula? I am looking for a way to solve $A^3+B^3$ the same way I would solve something like $(x^2 + 2)(x^2 - 2)$ by using the FOIL method to get the proper answer and not having to remember that it equals $A^2 - B^2$
September 1st, 2008, 06:32 PM   #5
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Re: The Sum of Cubes

Quote:
 Originally Posted by ruit Is there any way to get $(A + B)(A^2 - AB + B^2)$ without having to remember that specific formula?
Memorise it. You will meet it often in your travels, and it's easy enough if you stop trying to avoid it. You cn gain insight into it by looking at it form different perspectives:

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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