Manipulación de expansión binomial

$$\begin{align} &\;\sum_{j=0}^k \binom{k}{j} (-1)^j x^{k-j} \left( y^j - \sum_{i=0}^j \binom{j}{i} y^{j-i} z^i \right)\\ =&\;\sum_{j=0}^k \binom{k}{j} x^{k-j}(-y)^j - \sum_{j=0}^k\binom{k}{j} (-1)^j x^{k-j}\sum_{i=0}^j \binom{j}{i} y^{j-i} z^i \\ =&\;(x-y)^k - \sum_{j=0}^k\binom{k}{j} (-1)^j x^{k-j}(y+z)^j \\ =&\;(x-y)^k - \sum_{j=0}^k\binom{k}{j} x^{k-j}(-y-z)^j \\ =&\;(x-y)^k - (x-y-z)^k \\ \end{align}$$

Si quieres sacar un factor de$z$,

$$\begin{align} &\;(x-y)^k - (x-y-z)^k \\ =&\;(x-y)^k -\sum_{j=0}^k \binom{k}{j} ( x-y)^{k-j}(-z)^j \\ =&\;(x-y)^k - ( x-y)^{k}-\sum_{j=1}^k \binom{k}{j} ( x-y)^{k-j}(-z)^j \\ =&\;z\sum_{j=1}^k \binom{k}{j} ( x-y)^{k-j}(-z)^{j-1} \end{align}$$

Si el enfoque principal es la factorización$z$solamente, tenga en cuenta el sumando de la suma interna con$i=0$es igual a$y^j$.

Entonces,$y^j$se cancela y obtenemos

$$\begin{align*} \sum_{j=0}^k& \binom{k}{j} (-1)^j x^{k-j} \left( y^j - \sum_{i=0}^j \binom{j}{i} y^{j-i} z^i \right)\\ &=\sum_{j=0}^k \binom{k}{j} (-1)^{j+1} x^{k-j}\sum_{i=1}^j \binom{j}{i} y^{j-i} z^i\\ &=z\left(\sum_{j=0}^k \binom{k}{j} (-1)^{j+1} x^{k-j}\sum_{i=1}^j \binom{j}{i} y^{j-i} z^{i-1}\right)\\ &=zP(x,y,z) \end{align*}$$ con $P(x,y,z)$un polinomio en $x,y$y $z$.