2011/077) factor (x+y+z)^5 -x^5-y^5-z^5.?

Note
factorization is base don understanding symmetric and finding some method based on it

Method
we see that (x+y+z)^5 - x^5 - (y^5+z^5)

(x+y+z)^5 - x^5 is divisible by y+z and y^5+z^5 by y+ z

so it is divisible by (y+z)

hence it is divisible by (z+x) and (x+y) by symmetry so (x+y)(x+z)(y+z)

(x+y+z)^5- x^5-y^5-z^5

= x^5+ 5x^4(y+z) + 10 x^3(y+z)^2 + 10 x^2(y+z)^3 + 5x(y+z)^4+ (y+z)^5 - - x^ 5- y^5 - z^5
= 5x^4(y+z) + 10 x^3(y+z)^2 + 10 x^2(y+z)^3 + 5x(y+z)^4 + (y+z)^5- (y^5+z^5)

Now (y+z)^5 = y^5+ 5y^4z + 10 y^3z^2 + 10y^2z^3+5yz^4 + z^5

Hence (y+z)^5- y^5-z^5 = 5y^4z + 10 y^3z^2 + 10y^2z^3+5y^4

= 5y^4z + 5 zy^4 + 10 y^3z^2 + 10y^2z^3
= 5yz(y^3+z^3) + 10 y^2z^2(y+z)
= 5yz(y+z)(y^2-yz+z^3) + 10 y^2z^2(y+z)
= 5yz[(y+z)(y^2-yz+z^2) + 2yz)
= 5yz(y+z)(y^2+yz + z^2)

So 5x^4(y+z) + 10 x^3(y+z)^2 + 10 x^2(y+z)^3 + 5x(y+z)^4 + (y+z)^5- (y^5+z^5)
= 5x^4(y+z) + 10 x^3(y+z)^2 + 10 x^2(y+z)^3 + 5x(y+z)^4 + 5yz(y+z)(y+yz+z^2)
= 5(y+z) (x^4+ 2x^3(y+z) + 2x^2(y+z)^2 + x(y+z)^3 + yz(y^2+z^2+ yz))

Now we need to factor

(x^4+ 2x^3(y+z) + 2x^2(y+z)^2 + x(y+z)^3 + yz(y^2+z^2+yz) …1

This has a factor x + y and also x+ z so we have a factor (x^2+(y+z)x + yz) so let other factor be (x^2+ax+c)

(x^2+(y+z)x + yz) (x^2+ax+c)

= x^4 + x^3(a+y+z) + x^2(c+a(y+z)+ yz) + x(ayz+c(y+z)) + cyz) …2

Comparing coefficients from 1 and 2
From x^3
a= y+z,
from constant
c= y^2+z^2+yz

so coefficient of x^2 in 2nd expression that is (2) = c+ a(y+z) + yz
= (y^2+z^2 + yz + (y+z)^2 + yz)
= (y^2 + z^2 + 2yz) + (y+z)^2 = 2(y2+z)^2 matches in (1)
Coefficient of x (in 2)
= ayz + c(y+z)
= (y+z)yz + (y^2+z^2+yz) (y+z) = (y+z)^3

So the coefficient of x^2 and x match in 1 and 2 and hence the value is consistent

So we have
(x+y+z)^5 - x^5 - (y^5+z^5)
= 5(y+z) (x^4+ 2x^3(y+z) + 2x^2(y+z)^2 + x(y+z)^3 + yz(y^2+z^2+ yz))
= 5(y+z)(x+z)(x+y)(x^2+ yx + zx + y^2+z^2+ yz)

This is the complete factorization