- #1

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(A*B + C*D + E*F)

^{2}

Is there any kind of shortcut (even a website that does it automatically!) that can help me solve expressions like this?

- Thread starter Dawei
- Start date

- #1

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(A*B + C*D + E*F)

Is there any kind of shortcut (even a website that does it automatically!) that can help me solve expressions like this?

- #2

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I'm not entirely sure what you mean by "simplify" because it already seems to be in simplified form. Did you mean expand? If you meant expand..

Then use the general [tex] {(a +b)}^2 = a^2 +2ab +b^2 ...

let AB+CD =a, and EF = b

[/tex]

- #3

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http://www.flickr.com/photos/41988307@N08/4928253310/sizes/l/in/photostream/

- #4

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I mean... there isn't really any shortcut to this (If thats what you are looking for)

http://www.flickr.com/photos/41988307@N08/4928253310/sizes/l/in/photostream/

You can do (A*B + C*D + E*F)^2 = (A*B + C*D + E*F)*(A*B + C*D + E*F) and just multiply it out term by term.. or you can let a = AB+CD, and b = EF and use the general form for [tex] (a+b)^2 [/tex]

[tex] so.. (AB + CD)^2 +2EF(AB +CD) + E^{2}F^2 = A^2B^2 + C^2D^2 +E^2F^2 + 2(ABCD + EFAB + EFCD) [/tex]

I hope that's what you were looking for...or i'm not getting your question

- #5

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Thanks again.

- #6

CRGreathouse

Science Advisor

Homework Helper

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http://www.wolframalpha.com/

which is pretty poor overall but should be able to do what you want here.

- #7

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There is a trinominal formula, which can be generalized: (a+b+c)^3 = (3!/3!)[a^3+b^3+c^3) + 3!/2!1! [a^2(b+c)+b^2[a+c]+c^2[a+b]+3!/1!1!1! (abc).

This can be applied to (a+b+c)^2 since there are only two kinds of answers m^2 or mn. The coefficients are 2!/2! or 2!/1!1!, which is 2 or 1, of course. We just rely on symmetry. a^2+b^2+c^2+[2](ab+ac+bc)].

So this is rather easily handled by symmetry. More difficult cases can be handled, such as (a+b+c+d)^4. The coefficients are going to be like 4!/4! = 1, 4!/3!1! = 4, 4!/2!2! = 6, 4!/2!1!1!=12, and 4!/1!1!1!1! =24, the last coefficient being used in 24abcd.

This can be applied to (a+b+c)^2 since there are only two kinds of answers m^2 or mn. The coefficients are 2!/2! or 2!/1!1!, which is 2 or 1, of course. We just rely on symmetry. a^2+b^2+c^2+[2](ab+ac+bc)].

So this is rather easily handled by symmetry. More difficult cases can be handled, such as (a+b+c+d)^4. The coefficients are going to be like 4!/4! = 1, 4!/3!1! = 4, 4!/2!2! = 6, 4!/2!1!1!=12, and 4!/1!1!1!1! =24, the last coefficient being used in 24abcd.

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