Question

prove a^3 +b^3 +c^3 -3abc= (a+b+c)(a^2 +b^2 +c^2 -ab-bc-ca)

Answer 1

LHS

a^3 + b^3 + c^3 - 3abc 

factor a^3 + b^3 using cubic formula 

(a+b)(a^2 - ab + b^2) + c^3 - 3abc 

now we add 3ab and subtract 3ab at the same time into (a^2 -ab + b^2) get: 

(a+b)(a^2 +2ab +b^2 -3ab) + c^3 - 3abc 
*note that (a^2 + 2ab + b^2 -3ab) = (a^2 - ab + b^2) 

Now that we have factors of a perfect square in (a^2 +2ab +b^2) we can compress it to form 

(a+b)[ (a+b)^2 -3ab)] + c^3 - 3abc 
all i did was change (a^2 +2ab +b^2 -3ab) to [(a+b)^2 -3ab)] 

distribute the (a+b) which gives us 
(a+b)^3 -3ab(a+b) + c^3 - 3abc 

Again we have a perfect cube so we can factor to get our (a + b + c) term. The two cubic terms are (a+b)^3 and c^3 
---note (a+b)^3 + c^3 = (a+b+c)[(a+b)^2-c(a+b) + c^2] 

(a+b+c)[(a+b)^2-c(a+b) + c^2] - 3ab(a+b) - 3abc

Now we factor out the -3ab on the right hand side to get 

(a+b+c)[(a+b)^2-c(a+b) + c^2] - 3ab(a+b+c) 

Then we factor out the (a+b+c) 

(a+b+c)[(a+b)^2-c(a+b) + c^2 - 3ab] 

the term [(a+b)^2-c(a+b) + c^2 - 3ab] turns into (a^2 + b^2 + c^2 - ab - ac - bc) when fully factored and the terms canceled, and thus we get 

(a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc)