Question

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

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

(a+b+c)³=[a+(b+c)]³

=a³+(b+c)³+3a(b+c)[a+(b+c)]

=a³+b³+c³+3bc(b+c)+3a(b+c)[a+(b+c)]

=a³+b³+c³+3(b+c)[bc+a{a+(b+c)}]

=a³+b³+c³+3(b+c)[bc+a²+ab+ac]

=a³+b³+c³+3(b+c)[a(a+b)+c(a+b)]

=a³+b³+c³+3(b+c)(a+c)(a+b)

(a+b+c)³-a³-b³-c³=3(a+b)(b+c)(c+a)

hence, proved