a^3+b^3=(a+b)(a^2-ab+b^2 )
Sub b=b+c
a^3+(b+c)^3=(a+b+c)(a^2-a(b+c)+(b+c)^2 )
a^3+b^3+〖3b〗^2 c+〖3bc〗^2+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )
a^3+b^3+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2
a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2-3abc
a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-3bc(a+b+c)
a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2-3bc)
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
This is one method of getting that hope you understand...
on thing to notice about is that it is symmetric in a,b and c: switch any two of a,b, and c and you still get the same thing.
so if it factors, it will factor as two symmetric expressions, one involving just a single term at a time, and one involving two terms at a time.
but the only symmetric expression in a,b,c involving just a single term at a time is: a+b+c, or some multiple thereof. so let's divide by a+b+c,
considering it as a polynomial in a.
obviously the first term of the quotient is going to be , so we get:
. subtracting that from gives us:
. the term's coefficient is , so we take:
, and subtract THAT from our previous remainder:
.
now the coefficient of a is , so we take:
.
as this is equal to our last remainder, we have:
or, re-arranging a little:
.