>>454
(a+b+c)^3-a^3-b^3-c^3 =(a+b+c)^3-a^3-(b+c)^3+3bc(b+c)
=(b+c){(a+b+c)^2+(a+b+c)a+a^2}-(b+c){(b+c)^2-3bc}
=(b+c){3a^2+3a(b+c)+(b+c)^2}-(b+c){(b+c)^2-3bc}
=3(b+c){a^2+(b+c)a+bc}
=3(a+b)(b+c)(c+a)