Question

Prove that (a+b+c)^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a)

Answer 1

Proof + Step by step explanation:

LHS:

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

={(a+b)+c}^3- a^3-b^3-c^3

=(a+b)^3+c^3+3c(a+b)(a+b+c)-a^3-b^3-c^3 =a^3+b^3+c^3+3ab(a+b)+3c(a+b)(a+b+c)-a^3-b^3-c^3=3ab(a+b)+3c(a+b)(a+b+c)=3(a+b){ab+c(a+b+c)}=3(a+b){ab+ac+bc+c^3}=3(a+b){a(b+c)+c(b+c)} =3(a+b)(b+c)(c+a)

Hence proved

Answer 2

Step-by-step explanation:

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

=((a+b)+c}^3- a^3-b^3-c^3

=(a+b)^3+c^3+3c(a+b)(a+b+c)-a^3-b^3-c^3 =a^3+b^3+c^3+3ab(a+b)+3c(a+(a+b+c)-a^3-b^3-c^3

=3ab(a+b)+3c(a+b) (a+b+c)=3(a+b){ab+c(a+b+c)}

=3(a+b) {ab+ac+bc+c^3}=3(a+b){a(b+c)+c(b+c)}

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

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