Determine the set of real numbers $\alpha$ that can be expressed in the form \[\alpha=\sum_{n=0}^{\infty}\frac{x_{n+1}}{x_n^3}\]
where $x_0,x_1,x_2,\dots$  is an increasing sequence of real numbers with $x_0=1$.
To determine the set of real numbers $\alpha$ that can be expressed in the form 
\[
\alpha = \sum_{n=0}^{\infty} \frac{x_{n+1}}{x_n^3}
\]
where $x_0, x_1, x_2, \dots$ is an increasing sequence of real numbers with $x_0 = 1$, we consider two cases: when the sequence $\{x_n\}$ is bounded and when it is unbounded.

1. **Case 1: $\{x_n\}$ is bounded**

   Suppose $\{x_n\}$ is bounded, i.e., there exists some $M > 0$ such that $x_n \le M$ for all $n \in \mathbb{N}$. Then, we have:
   \[
   \frac{x_{n+1}}{x_n^3} \ge \frac{1}{x_n^2} \ge \frac{1}{M^2}
   \]
   Since $\sum_{n=0}^{\infty} \frac{1}{M^2}$ is a divergent series (as it is a constant positive term summed infinitely), it follows that:
   \[
   \sum_{n=0}^{\infty} \frac{x_{n+1}}{x_n^3}
   \]
   is also divergent. This leads to a contradiction because $\alpha$ must be finite. Therefore, $\{x_n\}$ cannot be bounded.

2. **Case 2: $\{x_n\}$ is unbounded**

   Since $\{x_n\}$ is increasing and unbounded, we have $\lim_{n \to \infty} x_n = \infty$. We need to estimate the sum:
   \[
   \sum_{n=0}^{\infty} \frac{x_{n+1}}{x_n^3}
   \]
   We use the inequality:
   \[
   \frac{x_{n+1}}{x_n^3} \ge \frac{3\sqrt{3}}{2} \left( \frac{1}{x_n^2} - \frac{1}{x_{n+1}^2} \right)
   \]
   This inequality is equivalent to:
   \[
   (x_{n+1} - \sqrt{3} x_n)^2 (x_{n+1} + \frac{\sqrt{3}}{2} x_n) \ge 0
   \]
   which is clearly true since the left-hand side is a square term multiplied by a positive term.

   Summing both sides from $n=0$ to $\infty$, and noting that $\lim_{n \to \infty} x_n = \infty$, we get:
   \[
   \alpha = \sum_{n=0}^{\infty} \frac{x_{n+1}}{x_n^3} \ge \sum_{n=0}^{\infty} \frac{3\sqrt{3}}{2} \left( \frac{1}{x_n^2} - \frac{1}{x_{n+1}^2} \right)
   \]
   The series on the right-hand side is a telescoping series, which simplifies to:
   \[
   \sum_{n=0}^{\infty} \left( \frac{1}{x_n^2} - \frac{1}{x_{n+1}^2} \right) = \frac{1}{x_0^2} = 1
   \]
   Therefore:
   \[
   \alpha \ge \frac{3\sqrt{3}}{2} \cdot 1 = \frac{3\sqrt{3}}{2}
   \]

   For the construction, let $\alpha \ge \frac{3\sqrt{3}}{2}$. Then there exists a $t \ge \sqrt{3}$ such that:
   \[
   t + \frac{3\sqrt{3}}{2} \cdot \frac{1}{t^2} = \alpha
   \]
   Choosing $x_n = (\sqrt{3})^{n-1} \cdot t$ for $n \ge 1$ suffices.

The final answer is $\boxed{\alpha \ge \frac{3\sqrt{3}}{2}}$