Question 1020445
<pre>
The other tutor gave a partial factorization but not 
the complete factorization.  Let's go for the complete 
factorization, not just show that we can factor out 3.

{{{(x+y+z)^3 - x^3 - y^3 - z^3}}}

This is a third degree polynomial in three variables.

Set it equal to 0, and look for its zeros:

{{{(x+y+z)^3 - x^3 - y^3 - z^3}}}{{{""=""}}}{{{0}}}

If we assume x=-y we get

{{{(-y+y+z)^3 - (-y)^3 - y^3 - z^3}}}{{{""=""}}}{{{0}}}

{{{z^3 + y^3 - y^3 - z^3}}}{{{""=""}}}{{{0}}}

{{{0}}}{{{""=""}}}{{{0}}}

So since x=-y  gives an identity, that means that

(x+y) is a factor of the given polynomial.

In exactly the same way, by symmetry x=-z and y=-z will 
also give an identity.

Therefore (x+y)(x+z)(y+z) must be a factor of the original
polynomial.

Since this will yield a third degree polynomial when
multiplied out, it can only be different from the factorization
of the original polynomial by a non-zero constant factor.

So the factorization must be:

{{{k(x+y)(x+z)(y+z)}}}, 

for some non-zero constant k. So

{{{(x+y+z)^3 - x^3 - y^3 - z^3}}}{{{""=""}}}{{{k(x+y)(x+z)(y+z)}}}

must be an identity for all values of x,y, and z

Let's choose x = 1, y = 1, z = 0

{{{(1+1+0)^3 - 1^3 - 1^3 - 0^3}}}{{{""=""}}}{{{k(1+1)(1+0)(1+0)}}}

{{{2^3 - 1 - 1}}}{{{""=""}}}{{{k(2)(1)(1)}}}

{{{8-2}}}{{{""=""}}}{{{2k}}}

{{{6}}}{{{""=""}}}{{{2k}}}

{{{3}}}{{{""=""}}}{{{k}}}

Therefore the factorization {{{k(x+y)(x+z)(y+z)}}}

becomes

{{{3(x+y)(x+z)(y+z)}}}

That took longer than 30 seconds!  Sorry! But we got it done :)

Edwin</pre>