Binomial expansion manipulation

Question

Can someone help me manipulate the following binomial expansion such that I can pull a factor of $z$ out of $$\sum_{j=0}^k \binom{k}{j} (-1)^j x^{k-j} \left( y^j - \sum_{i=0}^j \binom{j}{i} y^{j-i} z^i \right)$$

If you put this into Wolfram, every factor has a $z$ in it, but I can't figure out how to manipulate the above expression such that I can write $$z \left( \sum_{j=0}^k \cdots \right)$$

Answer 1

\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}

If you want to pull out a factor of $z$,

\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}

Answer 2

If the main focus is factoring out $z$ only, note the summand of the inner sum with $i=0$ is equal to $y^j$.

So, $y^j$ cancels out and we obtain \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*} with $P(x,y,z)$ a polynomial in $x,y$ and $z$.