How on earth will anyone prove $n^3-3n^2+n-1=Θ(n^3)$ I know its homework question.Sorry for that.But i was solving all problems of Skiena and chapter and got stuck to this problem of 2nd chapter 2.10.
Its easy to prove $n^3-3n^2+n-1=O(n^3)$ because $n^3-3n^2+n-1<=c.n^3$ for any $c>=1$
but how $n^3-3n^2+n-1=Ω(n^3)$ is possible.How to show $n^3-3n^2+n-1>=c.n^3$ for all values of $c>=1$ ?

Hint :Use limits to show that the quotient approaches one . Then use the definition of limit to show existence of the required constants 
