SOLUTION: Factor {{{(x+y+z)^3 - x^3 - y^3 - z^3}}} within 30 seconds?

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor {{{(x+y+z)^3 - x^3 - y^3 - z^3}}} within 30 seconds?      Log On


   



Question 1020445: Factor %28x%2By%2Bz%29%5E3+-+x%5E3+-+y%5E3+-+z%5E3 within 30 seconds?
Found 2 solutions by fractalier, Edwin McCravy:
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2By%2Bz%29%5E3+-+x%5E3+-+y%5E3+-+z%5E3=
=
3x%5E2y%2B3x%5E2z%2B3xy%5E2%2B6xyz%2B3xz%5E2%2B3y%5E2z%2B3yz%5E2=
3%28x%5E2y%2Bx%5E2z%2Bxy%5E2%2B2xyz%2Bxz%5E2%2By%5E2z%2Byz%5E2%29=
3%28x%5E2%28y%2Bz%29+%2B+y%5E2%28x%2Bz%29+%2B+z%5E2%28x%2By%29+%2B+2xyz%29
This can be further factored to
3%28x%2By%29%28x%2Bz%29%28y%2Bz%29

Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
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.

%28x%2By%2Bz%29%5E3+-+x%5E3+-+y%5E3+-+z%5E3

This is a third degree polynomial in three variables.

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

%28x%2By%2Bz%29%5E3+-+x%5E3+-+y%5E3+-+z%5E3%22%22=%22%220

If we assume x=-y we get

%28-y%2By%2Bz%29%5E3+-+%28-y%29%5E3+-+y%5E3+-+z%5E3%22%22=%22%220

z%5E3+%2B+y%5E3+-+y%5E3+-+z%5E3%22%22=%22%220

0%22%22=%22%220

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%28x%2By%29%28x%2Bz%29%28y%2Bz%29, 

for some non-zero constant k. So

%28x%2By%2Bz%29%5E3+-+x%5E3+-+y%5E3+-+z%5E3%22%22=%22%22k%28x%2By%29%28x%2Bz%29%28y%2Bz%29

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

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

%281%2B1%2B0%29%5E3+-+1%5E3+-+1%5E3+-+0%5E3%22%22=%22%22k%281%2B1%29%281%2B0%29%281%2B0%29

2%5E3+-+1+-+1%22%22=%22%22k%282%29%281%29%281%29

8-2%22%22=%22%222k

6%22%22=%22%222k

3%22%22=%22%22k

Therefore the factorization k%28x%2By%29%28x%2Bz%29%28y%2Bz%29

becomes

3%28x%2By%29%28x%2Bz%29%28y%2Bz%29

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

Edwin