Integration by Parts and Polynomial Identities (sum and difference of the n-th powers of x and y)

Consider $\begin{array}{c}\int u{v}^{(n+1)}dx=\int ud{v}^{(n)}=u{v}^{(n)}-\int u^{\prime }{v}^{(n)}dx\end{array}$$\int u v^{(n+1)}dx=\int ud v^{(n)}=u v^{(n)}-\int u' v^{(n)}dx\\$

Thus we have $\begin{array}{c}\int u^{\prime }{v}^{(n)}dx=\int u^{\prime }d{v}^{(n-1)}=u^{\prime }{v}^{(n-1)}-\int u^{″}{v}^{(n-1)}dx\end{array}$$\int u'v^{(n)}dx=\int u'd v^{(n-1)}=u' v^{(n-1)}-\int u'' v^{(n-1)}dx\\$

Therefore $\begin{array}{c}\int u{v}^{n+1}dx=u{v}^{(n)}-u^{\prime }{v}^{(n-1)}+u^{″}{v}^{(n-2)}+...+(-1{)}^n{u}^{(n)}v+(-1{)}^{n+1}\int u^{(n+1)}vdx\end{array}$$\int u v^{n+1}dx=uv^{(n)}-u' v^{(n-1)}+u'' v^{(n-2)}+...+(-1)^nu^{(n)}v+(-1)^{n+1}\int u^{(n+1)}vdx\\$

$\begin{array}{c}\int u{v}^{n+1}dx++(-1{)}^n\int u^{(n+1)}vdx=u{v}^{(n)}-u^{\prime }{v}^{(n-1)}+u^{″}{v}^{(n-2)}+...+(-1{)}^n{u}^{(n)}v\end{array}$$\int u v^{n+1}dx++(-1)^{n}\int u^{(n+1)}vdx=uv^{(n)}-u' v^{(n-1)}+u'' v^{(n-2)}+...+(-1)^nu^{(n)}v\\$

Derivative both sides, we have

$\begin{array}{c}u{v}^{(n+1)}+(-1{)}^n{u}^{(n+1)}v=\frac{d}{dx}(u{v}^{(n)}-u^{\prime }{v}^{(n-1)}+u^{″}{v}^{(n-2)}+...+u^{(n)}v)\end{array}$$u v^{(n+1)}+(-1)^{n} u^{(n+1)} v=\frac{d}{dx}(uv^{(n)}-u' v^{(n-1)}+u'' v^{(n-2)}+...+u^{(n)}v)\\$

We already know that $\begin{array}{c}\frac{d}{dx}(uv)=({\partial }_u+{\partial }_v)(uv)\end{array}$$\frac{d}{dx}(uv)=(\partial_u+\partial_v)(uv)\\$

And Consider the differential operator two sides,

${\partial }_v^{(n+1)}+(-1{)}^n{\partial }_u^{(n+1)}=({\partial }_v+{\partial }_u)({\partial }_v^{(n)}-{\partial }_u{\partial }_v^{(n-1)}+{\partial }_u^{(2)}{\partial }_v^{(n-2)}+...+(-1{)}^n{\partial }_u^{(n)})$$\partial_v^{(n+1)}+(-1)^n\partial_ u^{(n+1)}=(\partial_v+\partial_u)(\partial_v^{(n)}-\partial_u\partial_v^{(n-1)}+\partial_u^{(2)}\partial_v^{(n-2)}+...+(-1)^n\partial_u^{(n)})$

It's just polynomials identity $\begin{array}{c}{x}^{n+1}+(-1{)}^n{y}^{n+1}=(x+y)(x^n-x^{n-1}y+...+(-1{)}^{n-1}(x{y}^{n-1})+(-1{)}^n{y}^n)\end{array}$$x^{n+1}+(-1)^n y^{n+1}=(x+y)(x^n-x^{n-1}y+...+(-1)^{n-1}(xy^{n-1})+(-1)^n y^n)\\$

When $n=1$$n=1$, we have $x^2-y^2=(x+y)(x-y)$$x^2-y^2=(x+y)(x-y)$

When $n=2$$n=2$, we have $x^3+y^3=(x+y)(x^2-xy+y^2)$$x^3+y^3=(x+y)(x^2-xy+y^2)$

we can let $y:=-y$$y:=-y$ and get $x^{n+1}-y^{n+1}=(x-y)(x^n+x^{n-1}y+x^{n-2}{y}^2+...+y^n)$$x^{n+1}-y^{n+1}=(x-y)(x^n+x^{n-1} y+x^{n-2} y^2+...+y^n)$

Thus we have $x^{n+1}-y^{n+1}=(x-y)(x^n+x^{n-1}y+x^{n-2}{y}^2+...+y^n)$$x^{n+1}-y^{n+1}=(x-y)(x^n+x^{n-1} y+x^{n-2} y^2+...+y^n)$, for all $n\in \mathbb{N}$$n\in\mathbb N$

And $x^{n+1}+y^{n+1}=(x+y)(x^n-x^{n-1}y+...+(-1{)}^{n-1}(x{y}^{n-1})+y^n)$$x^{n+1}+y^{n+1}=(x+y)(x^n-x^{n-1}y+...+(-1)^{n-1}(xy^{n-1})+ y^n)$, for all $n\in 2\mathbb{N}$$n\in 2\mathbb N$

It shows an interesting way to derivate the identity