Question

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

Answer 1

Step-by-step explanation:

a+b+c)³-a³-b³-c³

=(a+b)³+3(a+b)²c+3(a+b)c²+c³-a³-b³-c³

=a³+3a²b+3ab²+b³+3(a²+2ab+b²)c+3ac²+3bc²+c³-a³-b³-c³

=3a²b+3ab²+3a²c+6abc+3b²c+3ac²+3bc²

=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)

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

=3(ab+b²+ac+bc)(c+a)

=3(abc+b²c+ac²+bc²+a²b+ab²+a²c+abc)

=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)

∴, LHS=RHS (Proved)

Answer 2

See the attachment. Hope, it'll help you.....