Factorization of vectors $(Z^T Z)^n Z^T-(X^T X)^n X^T$

Question

Is there a way to factorize expression $(Z^T Z)^k Z^T-(X^T X)^k X^T$ where $Z$ and $X$ are real column vectors in $\mathbb{R}^n$, such that \begin{align} (Z^T Z)^k Z^T-(X^T X)^k X^T= (Z-X)^T P(Z,X) \end{align} where $P(Z,X)$ is a polynomial.

My question is motivated by the following difference formula for real $a$ and $b$ \begin{align} a^k-b^k= (a-b) P(a,b) \end{align} where $P(a,b)= \sum_{i=0}^{k-1} a^i b^{k-1-i}$

Answer 1

If $\lVert X\lVert=\lVert Z\lVert=a$, then your factorization holds trivially, with $P=a^{2k}$.

If $X$ and $Z$ are not collinear, then let $\lVert X\lVert=a$ and $\lVert Z\lVert=b$, with $a>0,b>0$.

Suppose further that $(Z^T Z)^k Z^T-(X^T X)^k X^T=c(Z-X)^T$ for some real number $c$.

That is, $b^{2k}Z^T-a^{2k}X^T=cZ^T-cX^T$

Since $X$ and $Z$ are not collinear, coordinates along $X^T$ and $Z^T$ of the above linear combination are respectively equal.

Hence, $c=a^{2k}=b^{2k}$.

Therefore, if $X$ and $Z$ are not collinear, your factorization holds iff $\lVert X\lVert=\lVert Z\lVert$.