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Solution
We can see that at n = 2, $$n^3-11n^2+32n-28=0$$ i.e. (n-2) is a factor of $$n^3-11n^2+32n-28$$
$$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$$
We can further factorize n^2-9n+14 as (n-2)(n-7).
$$n^3-11n^2+32n-28=(n-2)^2(n-7)$$
$$\Rightarrow$$ $$n^3-11n^2+32n-28>0$$
$$\Rightarrow$$ $$(n-2)^2(n-7)>0$$
Therefore, we can say that n-7>0
Hence, n$$_{min}$$ = 8
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