O(n^2) vs O (n(logn)^2)

Question

Is time complexity O(n^2) or O (n(logn)^2) better?

I know that when we simplify it, it becomes

O(n) vs O((logn)^2)

Answer 1

O(n^2) vs. O(n*log(n)^2)
<=> O(n) vs. O(log(n)^2) (divide by n)
<=> O(sqrt(n)) vs. O(log(n)) (square root)
<=> polynomial vs. logarithmic

Logarithmic wins.