For $|t|>1,$ is it possible to "factor" the difference term $x^t-y^t$ ? For integer $t$ we know how to do that, and with some twist, for rational numbers $t$ as well. My question is whether we can do the same for real-valued $t$ to obtain a term of the form $(x-y) \cdot u(x,y).$
Please don't provide the trivial way of $u(x,y)=\frac{x^t-y^t}{x-y}.$
I hope this question is not a duplicate (I tried to search and did not find one).
Presumably you want a closed formula for $\frac{x^t-y^t}{x-y}.$ I doubt you'll find a good one. The heart of the problem is that the function $z^t$ on the complex plane can not be made an analytic function at $0$ when $t$ is not a non-negative integer.
But you can get something, using the power series for $(1+u)^t,$ which is only true for $|u|<1.$
For $|u|<1,$ $$(1+u)^t=\sum_{i=0}^{\infty}\binom ti u^i.$$ Then if $0<y\leq x,$ $$\begin{align} x^t-y^t&=x^t(1-(y/x)^t)\\&=x^t\left(1-\left(1-\frac{x-y}{x}\right)^t\right)\\ &=x^t\sum_{i=1}^\infty (-1)^{i-1}\binom ti\left(\frac{x-y}{x}\right)^i \end{align} $$
So $$\frac{x^t-y^t}{x-y}=\sum_{j=0}^\infty (-1)^j\binom{t}{j+ 1}x^{t-j-1}(x-y)^j$$
When $t$ is a non-negative integer, each term is a polynomial because $\binom{t}{j+1}=0$ when $j>t-1.$
But it depends on the order of $x,y.$ If $y>x,$ we have to reverse the variables.