(a+b+c)³-a³-b³-c³=3(a+b)(b+c)(c+a)
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(a+b+c)³-a³-b³-c³=3(a+b)(b+c)(c+a)
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)
∴, LHS=RHS (Proved)
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