Prove the identity (x+y)^5-(5yx^2+5xy^2)(x^2+xy+y^2)=x^5+y^5.

We'll
begin by expanding the
binomial:

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

We'll
subtract x^5 and y^5 from expansion as well as from the right side and we'll
get:

5x^4*y+10x^3*y^2+10x^2*y^3+5y^4*x

We'll
combine the middle terms and the extremes and we'll factorize
them:

5xy(x^3 + y^3) +
10x^2*y^2(x+y)

We recognize inside the 1st brackets, a sum of
cubes

x^3 +y^3 = (x+y)(x^2 - xy +
y^2)

5xy(x+y)(x^2 - xy + y^2) +
10x^2*y^2(x+y)

We'll factorize by
5xy(x+y)

5xy(x+y)(x^2 - xy + y^2 +
2xy)

We'll combine like terms inside
brackets:

LHS = 5xy(x+y)(x^2 + xy + y^2)=5xy(x+y)(x^2 +
xy + y^2) = RHS

We notice that managing both sides,
we've get the same expression, so the identity (x+y)^5-(5yx^2+5xy^2)(x^2+xy+y^2)=x^5+y^5 is
verified.