Question

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

Answer 1

Here you go...

Answer:

(a+b+c)³-a³-b³-c³

=(a+b)³+c³+3(a+b)c(a+b+c)-a³-b³-c³

[Using the formula, (a+b)³=a³+b³+3ab(a+b)]

=a³+b³+3ab(a+b)+3c(a+b)(a+b+c)-a³-b³

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

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

=3(a+b)(ab+ca+cb+c²)

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

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

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