Prove that there exists a number $x$ such that $x^3 = 6$ I want to show that there exists a real number $x$ such that $x^3 = 6$. Here is what I have so far. $\\$

Let $S = \{x \mid x \in \mathbb{R}, x \geq 0, x^3 < 6\}$. By this definition, $S$ is nonempty since $0 \in S$, and also $S$ is bounded above since $2^3 = 8 > 6$. Thus, by the Completeness Axiom, $S$ has a least upper bound; call it $b$.  We will show that $b^3 = 6$ (and hence, there exists a real number such that $x^3 = 6$) by showing that we cannot have $b^3 > 6$ or $b^3 < 6$. 
First, for the sake of contradiction, suppose we had $b^3 > 6$. Then, we will show that we can choose a suitably small positive number $\epsilon$ such that $b - \epsilon$ is also an upper bound for $S$, which contradicts $b$ being the least upper bound. But, I'm not sure about how to find $\epsilon$. I tried expanding:
$$(b - \epsilon)^3 = b^3 - 3b^2\epsilon + 3b\epsilon^2 - \epsilon^3,$$
and from here, I think I'm supposed to use greater-than equalities to try and come up with $\epsilon$, but I'm not really sure how to do that. Any help is appreciated.

If you write it like so:
$$(b-\epsilon)^3 = b^3 - \epsilon(3b^2 - 3b\epsilon + \epsilon^2)$$
you can see that (if $\epsilon < \frac b2$, which can safely be assumed), you have $$(b-\epsilon)^3 = b^3 - \epsilon\cdot M$$
where $M>b$.
Now, look at the values of $b^3$ and $6$. There is a "gap" assumed to be between them, call it $\delta = b^3-6$. Now, clearly, $6=b^3-\delta$, but your number $b^3-\epsilon\cdot M$ can be made greater than $b^3-\delta$ if you pick a small enough $\epsilon$.
