Fourier Transform: Solving x_3(n) = (n-1)^2x(n)

[FrogPad]

I am stumped on this...

Given a discrete function, and transform pair: $$x(n) \leftrightarrow \hat x (e^{j\omega})$$

What is the transform of:
$$x_3(n) = (n-1)^2 x(n)$$


I really don't know how to do this. I have a table proprety for [itex]nx(n) [/tex], but nothing with $n^2 x(n)$. The only thing I can think of is expanding it as: $x_3(n) = (n-1)^2x(n) = n^2x(n) - 2nx(n) +x(n)$... but I'm stuck on the $n^2$ part. My intuition says that it has something to do with the differentiation property, but I'm really stuck, and can't figure this out. Any help would be awesome. thanks :)[/itex]

 

[FrogPad]

Easy. Got it finally.

Just write $$x_3(n) = ny(n)$$
where $$y(n) = nx(n)$$

and since I have the transform for $nx(n)$, it is cake.