Math, asked by anushkakulkarni33, 2 months ago

prove that (a+b+c)-a^3-b^3-c^3=3(ab)(b+c)(c+a)​

Answers

Answered by anku3842
0

Answer:

(a+b+c)

3

=[(a+b)+c]

3

=(a+b)

3

+3(a+b)

2

c+3(a+b)c

2

+c

3

=>(a+b+c)

3

=(a

3

+3a

2

b+3ab

2

+b

3

)+3(a

2

+2ab+b

2

)c+3(a+b)c

2

+c

3

=>(a+b+c)

3

=a

3

+b

3

+c

3

+3a

2

b+3a

2

c+3ab

2

+3b

2

c+3ac

2

+3bc

2

+6abc

=>(a+b+c)

3

=a

3

+b

3

+c

3

+3a

2

b+3a

2

c+3ab

2

+3b

2

c+3ac

2

+3bc

2

+3abc+3abc

=>(a+b+c)

3

=a

3

+b

3

+c

3

+3a(ab+ac+b

2

+bc)+3c(ab+ac+b

2

+bc)

=>(a+b+c)

3

=a

3

+b

3

+c

3

+3(a+c)(ab+ac+b

2

+bc)

=>(a+b+c)

3

=a

3

+b

3

+c

3

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

=>(a+b+c)

3

=a

3

+b

3

+c

3

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

=>(a+b+c)

3

−a

3

−b

3

−c

3

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

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