Question

prove that (a+b+c) cube - a cube - b cube - c cube = 3(a+b)(b+c)(c+a)

Answer 1

the HTML source of the page has the original problem: 
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a + b + c^2 - ab - bc - ca) 

I'm also going to assume that what you really meant to type was the following, because otherwise the statement doesn't hold: 
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) 

Using the distributive rule on the right, we get 
a(a^2 + b^2 + c^2 - ab - bc - ac) + 
b(a^2 + b^2 + c^2 - ab - bc - ac) + 
c(a^2 + b^2 + c^2 - ab - bc - ac). 

Applying it once again, we get: 
a^3 + ab^2 + ac^2 - (a^2)b - abc - (a^2)c + 
(a^2)b + b^3 + bc^2 - ab^2 - (b^2)c - abc + 
(a^2)c + (b^2)c + c^3 - abc - bc^2 - (c^2)a 

Combining like terms, this simplifies to: 
a^3 + b^3 + c^3 - 3abc