Given an analysis of an algorithm, can we get Big-O, Theta, and Omega from this analysis?
The key point is that $O(\cdot)$ and friends describe relationships between functions. Functions are purely mathematical objects and it doesn't matter what they're being used to measure. This is similar to the way that $&lt;$ and friends describe relationships between numbers and it doesn't matter what those numbers are being used to measure.

So, in particular, it doesn't matter if your function is being used to measure the running time of some algorithm. It doesn't even matter if your function is being used to give, say, an upper bound on the running time of an algorithm. If I want to, I can give you an upper bound on a function that's a lower bound for the running time.  It's not very useful to do that but, because $O(-)$ doesn't care what the function is measuring, it makes perfect mathematical sense.  This is a bit like saying, "Jane is less than 1.75m tall and John is taller than her": I've given you an upper bound for Jane's height and Jane's height is, in turn, being used as a lower bound for John's. OK, John could have any height at all, but the mathematics makes sense.

So, all $T(n)=O(f(n))$ says is that the function $g$ is an upper bound for the function $T$, up to constant factors, for large enough inputs. It says nothing at all about any connection between the function $T$ and the real world: it doesn't say that $T$ is an upper bound for this or a lower bound for that.

With all of that out of the way, we can address the technical part of the question, but that really does just follow from the definitions, as I stated in comments.

Tou have a function $5n^3 + 100n^2 + 32n$ and you want to compare it to some other functions: $n$, $n^2$ and $n^3$. Let's call that function $T(n)$.  Apparently, the function&nbsp;$T$ is somehow related to the running time of some algorithm but that doesn't matter for what's about to happen.


$T(n) = O(n^3)$ because, for $n&gt;100$, $T(n) \leq 5n^3 + n\cdot n^2 + n^2\cdot n = 7n^3$.
$T(n)\neq O(n^2)$ because, for any constant&nbsp;$c$, and any $n\geq c$, $5n^3 + 100n^2 + 32n &gt; 5n^3 \geq 5cn^2 &gt; cn^2$. Similarly, $T(n)\neq O(n)$.
$T(n) = \Omega(n^3)$ since, for any $n&gt;0$, $T(n)&gt; n^3$. Similarly, $T(n) = \Omega(n^2)$ and $T(n) = \Omega(n)$.
$T(n) = \Theta(n^3)$ because $T(n) = O(n^3)$ and $T(n) = \Omega(n^3)$. $T(n)\neq \Theta(n^2)$ because $T(n)\neq O(n^2)$ and $T(n)\neq \Theta(n)$ because $T(n)\neq O(n)$.