SOLUTION: multiply using formula :(a+b+c) (b+c-a) (c+a-b) (a+b-c)

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Question 1043966: multiply using formula :(a+b+c) (b+c-a) (c+a-b) (a+b-c)
Answer by josgarithmetic(39627) About Me  (Show Source):
You can put this solution on YOUR website!
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Several use of number property arrangements allows the advantage of Difference of Two Squares.
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%28a%2Bb%2Bc%29%28a%2Bb-c%29%28-a%2Bb%2Bc%29%28a-b%2Bc%29
%28a%2Bb%2Bc%29%28a%2Bb-c%29%28-1%29%28a-b-c%29%28a-b%2Bc%29
%28%28a%2Bb%29%2Bc%29%28%28a%2Bb%29-c%29%28-1%29%28%28a-b%29-c%29%28%28a-b%29%2Bc%29
%28%28a%2Bb%29%5E2-c%5E2%29%28-1%29%28%28a-b%29%5E2-c%5E2%29

%28a%5E2%2B2ab%2Bb%5E2-c%5E2%29%28-1%29%28a%5E2-2ab%2Bb%5E2-c%5E2%29
%28-1%29%28a%5E2%2B2ab%2Bb%5E2-c%5E2%29%28a%5E2-2ab%2Bb%5E2-c%5E2%29-----Either this one or
%28-1%29%28a%5E2%2Bb%5E2%2B2ab-c%5E2%29%28a%5E2%2Bb%5E2-2ab-c%5E2%29-----this one would be best
done using a lattice arrangement for multiplying.

Sixteen terms initially occur and a few of them drop due to additive inverses.


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highlight%28-a%5E4-b%5E4-c%5E4%2B2a%5E2b%5E2%2B2a%5E2c%5E2%29-----Or this can be arranged for neatness of indicating the additions of terms.