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

Asked by anum82666 | 22nd Feb, 2020, 10:15: AM

Expert Answer:

(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

Answered by Arun | 23rd Feb, 2020, 11:14: AM