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.

Note that$$b^3 - 3b^2\varepsilon + 3b\varepsilon^2 - \varepsilon^3>6\iff b^3-6>3b^2\varepsilon-3b\varepsilon^2+\varepsilon^3.$$Now, take $\varepsilon\in\left(0,1\right)$ such that$$\varepsilon<\dfrac{b^3-6}{6b^2\varepsilon}\tag1$$and that$$\varepsilon<\dfrac{b^3-6}2.\tag2$$Then\begin{align}3b^2\varepsilon+\varepsilon^3&<3b^2\varepsilon+\varepsilon\text{ (because $\varepsilon<1$)}\\&<\frac{b^3-6}2+\frac{b^3-6}2\text{ (by $(1)$ and $(2)$)}\\&=b^3-6.\end{align}Therefore$$3b^2\varepsilon-3b\varepsilon^2+\varepsilon^3<3b^2\varepsilon+\varepsilon^3<b^3-6.$$
