We have to prove that (x+y)^5 - (5yx^2+5xy^2)(x^2+xy+y^2) = x^5 + y^5.

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

(5yx^2+5xy^2)(x^2+xy+y^2) = 5yx^4 + 5y^2x^3 + 5y^3x^2 + 5x^3y^2 + 5x^2y^3 + 5xy^4

Let's start from the left hand side

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

cancel the common terms

=> x^5 + y^5

which is the right hand side

**This proves that (x+y)^5 - (5yx^2+5xy^2)(x^2+xy+y^2) = x^5 + y^5.**

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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.**