Math, asked by op49942, 1 year ago

Prove that (a+b+c)3 - a3-b3-c3=3(a+b)(b+c)(c+a)

Answers

Answered by aqibkincsem
549
Proving the above is easy as long as you are well versed with some of the existing formula’s and understand how to break the pattern as we progress to obtain the desired result.

Please find below the solution for it:

 Consider:

(a+b+c)3 - a3-b3-c3 = [(a+b)+c]3 – a3-b3-c3

(a+b)3+c3+3c(a+b)(a+b+c)-a3-b3-c3

A3+b3+3ab(a+b)+c3+3c(a+b)(a+b+c)-a3-b3-c3

A3+b3+c3+3ab(a+b)+3c(a+b)(a+b+c)-a3-b3-c3

3ab(a+b)+3c(a+b)(a+b+c) 3(a+b)[ab+c(a+b+c)]

3(a+b)[ab+ac+bc+c2] 3(a+b)[a(b+c)+c(b+c)]

3(a+b)(b+c)(c+a)

 Hence Proved.
Answered by khushi7522
273

answer is.......

hope it is thankful...☺☺

Attachments:
Similar questions