Factorize a3+b3+c3-3abc.
Step 1: Factorize the given expression
a3+b3+c3-3abc
use the identity
(x+y)3=x3+y3+3xy(x+y)⇒(x+y)3-3xy(x+y)=x3+y3
put x=b,y=c
∴a3+b3+c3-3abc
=a3+(c+b)3-3cb(c+b)-3abc=a3+(b+c)3-3bc(a+b+c)
Step 2: Factorize further
using the identity x3+y3=(x+y)(x2+y2-xy)
x=a,y=b+c
a3+(b+c)3=(a+b+c)a2+(b+c)2-a(b+c)
∴a3+(b+c)3-3bc(a+b+c)=(a+b+c)a2+(b+c)2-a(b+c)-3bc(a+b+c)=(a+b+c)a2+b2+c2+2bc-ab-ac-3bc⇒a3+(b+c)3-3bc(a+b+c)=(a+b+c)a2+b2+c2-bc-ab-ac
Hence the factorized form of the polynomial a3+b3+c3-3abc is (a+b+c)(a2+b2+c2-bc-ab-ac).
The factors of a3+b3+c3−3abc are