Let $X$ be a topological space. Its Lebesgue dimension / topological covering dimension $dim(X)$ is defined as the minimal number $n$ for which, for every open cover $\{U\}$ of $X$ there is an open cover $\{V\}$ that refines $\{U\}$ such that every point $x\in X$ is contained in no more than $n+1$ sets of $\{V\}$. If no such minimal $n$ exists, $X$ is said to have infinite covering dimension.
Now, let $X$ be a locally compact Hausdorff space that is not compact, $X^\star$ be its one-point compactification, and $\beta X$ be its Stone-Cech compactification. It is direct to see that when $X=\mathbb{R}^n$, $dim(X)=dim(X^\star)$.
Question: Can one comment on $dim(X)$, $dim(X^\star)$, and $dim(\beta X)$ in the general case (as in, is there any ordering between them in general?)
I think there is another question and an answer that deals with somewhat related issues (for normal spaces), but that does not answer this question directly, I think.
$X$ is not a closed subspace of $X^\star$ or of $\beta X$, as it is not compact, so most of the usual theorems about closed subspaces having lower covering dimension, do not apply here (I think).
For any $T_4$ space $X$ $\dim(X)=\dim(\beta X)$ (e.g. see Engelking's dimension theory book, chapter 4; but I believe it's also in his "general topology" book), and it's clear also that $\dim(X^\ast)=\dim(X)$ for all locally compact Hausdorff spaces (addition theorems will show this e.g.)
Engelking later in chapter 4 shows also that a $T_4$ space has a compactfication of the same covering dimension such that the compactification has the same weight as well (which $\beta(X)$ will not have, in general).