The left-hand group has a factor of (x+y+z)-x=y+z; and the right-hand group also has a factor of y+z.  Then y+z is a factor of the whole expression.

We can do this two more ways, by pairing y^5 or z^5 with (x+y+z)^5 to see that x+z and x+y are also factors.

A bit of multiplication: (x+y) * (x+z) * (y+z) = x^2*y + x*y^2 + x^2*z + x*z^2 + y^2*z + y*z^2 + 2xyz

Then at this point we can say (x+y+z)^5 - x^5 - y^5 - z^5 

= (x^2*y + x*y^2 + x^2*z + x*z^2 + y^2*z + y*z^2 + 2xyz) * (Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz)

Then to calculate the coefficient A, we can look just at the x^4*y terms.  On the left side the coefficient of x^4*y can be found by evaluating the trinomial coefficient 5!/(4!*1!*0!) = 5.  On the right side, expanding the product will have a single x^4*y term from (x^2*y) * (Ax^2).  Then A=5.

Very similarly, B=5 and C=5, from looking at y^4*z and x*z^4.

So now we have (x+y+z)^5 - x^5 - y^5 - z^5 

= (x^2*y + x*y^2 + x^2*z + x*z^2 + y^2*z + y*z^2 + 2xyz) * (5x^2 + 5y^2 + 5z^2 + Dxy + Exz + Fyz)

This time we'll look at x^3*y^2 terms.  On the left side the trinomial coefficient is 5!/(3!*2!*0!) = 10.  And on the right side we have two terms: (x*y^2)*(5x^2) + (x^2*y)*(Dxy).  Then 10 = 5+D, so D=5.

Very similarly, E=5 and F=5, from looking at y^3*z^2 and x^2*z^3.

So gathering everything together, and pulling out a common constant factor, (x+y+z)^5 - x^5 - y^5 - z^5 = 5 * (x+y) * (x+z) * (y+z) * (x^2+y^2+z^2+xy+xz+yz).