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

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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 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 :)
 
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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.
 
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