Question

How many combination possible in a+b+c cube?

Answer 1

My reasoning:

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

(a+b)3+3(a+b)2c+3(a+b)c2+c3(a+b+c)3=[(a+b)+c]3=(a+b)3+3(a+b)2c+3(a+b)c2+c3(a+b+c)3=[(a+b)+c]3=(a+b)3+3(a+b)2c+3(a+b)c2+c3

(a+b+c)3=(a3+3a2b+3ab2+b3)+3(a2+2ab+b2)c+3(a+b)c2+c3(a+b+c)3=(a3+3a2b+3ab2+b3)+3(a2+2ab+b2)c+3(a+b)c2+c3(a+b+c)3=(a3+3a2b+3ab2+b3)+3(a2+2ab+b2)c+3(a+b)c2+c3

(a+b+c)3=a3+b3+c3+3a2b+3a2c+3ab2+3b2c+3ac2+3bc2+6abc(a+b+c)3=a3+b3+c3+3a2b+3a2c+3ab2+3b2c+3ac2+3bc2+6abc(a+b+c)3=a3+b3+c3+3a2b+3a2c+3ab2+3b2c+3ac2+3bc2+6abc

(a+b+c)3=(a3+b3+c3)+(3a2b+3a2c+3abc)+(3ab2+3b2c+3abc)+(3ac2+3bc2+3abc)−3abc(a+b+c)3=(a3+b3+c3)+(3a2b+3a2c+3abc)+(3ab2+3b2c+3abc)+(3ac2+3bc2+3abc)−3abc(a+b+c)3=(a3+b3+c3)+(3a2b+3a2c+3abc)+(3ab2+3b2c+3abc)+(3ac2+3bc2+3abc)−3abc

(a+b+c)3=(a3+b3+c3)+3a(ab+ac+bc)+3b(ab+bc+ac)+3c(ac+bc+ab)−3abc(a+b+c)3=(a3+b3+c3)+3a(ab+ac+bc)+3b(ab+bc+ac)+3c(ac+bc+ab)−3abc(a+b+c)3=(a3+b3+c3)+3a(ab+ac+bc)+3b(ab+bc+ac)+3c(ac+bc+ab)−3abc

(a+b+c)3=(a3+b3+c3)+3(a+b+c)(ab+ac+bc)−3abc(a+b+c)3=(a3+b3+c3)+3(a+b+c)(ab+ac+bc)−3abc(a+b+c)3=(a3+b3+c3)+3(a+b+c)(ab+ac+bc)−3abc

(a+b+c)3=(a3+b3+c3)+3[(a+b+c)(ab+ac+bc)−abc](a+b+c)3=(a3+b3+c3)+3[(a+b+c)(ab+ac+bc)−abc]