Question

1. Prove that: $(a+b+c)^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a)$

Answer 1

take the first term
(a+b+c)^3 - a^3-b^3-c^3
take (a+b+c)^3 write it as ((a+b)+c)^3 which is in the form of (x+y)^3.
expand it. you will get the answer as a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3bc^2+3b^2c+6abc
since you need to minus a^3-b^3-c^3
minus them.
answer is .: 3a^2b+3ab^2+3a^2c+3ac^2+3bc^2+3b^2c+6abc

then take the next term : that is 3(a+b)(b+c)(c+a)
first multiply (a+b)(b+c). then multiply its result with (c+a) you will get the answer as : 2abc+ac^2+a^2c+bc^2+b^2c+a^2b+b^2a

since, this is (a+b)(b+c)(c+a) you need to multiply it with 3.
then the answer is 6abc+3ac^2+3a^2c+3bc^2+3b^2c+3a^2b+3b^2a

then the lhs=rhs.

thank you.