Math, asked by vershamallick29, 5 months ago

using formula lets multiply: (a+b+c) (b+c-a) (c+a-b) (a+b-c)​

Answers

Answered by vansh1873
2

Answer:

(a+b+c)(c+a-b)(a+b-c)

(a-a)(b-b)(c-c)a+b+c

a+b+c

Answered by richashah042
1

Answer:

%28a%2Bb%2Bc%29%28b%2Bc-a%29%28c%2Ba-b%29%28a%2Bb-c%29

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.

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