Prove that for any real number $x$ there exists a real number $y$ with $y^3 = x$ There is no assumption that $x$ or $y$ are positive. It looks like this problem should be divided into two cases, one where $x \geq 0$ and the other $x < 0$.
 A: The function $f : x \mapsto x^3$ is continuous (prove this! One way is to use the fact that the product of continuous functions is continuous, and $x \mapsto x$ is continuous.) Furthermore, $n^3 \ge n$, and by the Archimedean property, $\mathbb{N}$ is unbounded in $\mathbb{R}$. Hence $f$ is a continuous unbounded function with $f(0) = 0$, so by the intermediate value theorem, every positive number has a cube root. Likewise, $-n^3 \le -n$, so we can do something similar for negative numbers.

However, I'm guessing that this is not at all the style proof you're looking for; I'd guess that you're considering the completeness of $\mathbb{R}$ and trying to use it; if so, think about ($x > 0)$ the set
$$A = \{y \in \mathbb{R} : y^3 < x\}$$
Then prove that $\sup A$ is the desired object.

Either method is a more-or-less direct application of completeness (that's the key ingredient in proving the IVT, after all).
A: Another idea is to identify the function $y:x\rightarrow x^3$ as the composition of two functions one of which is $y_1:x\rightarrow x^2$ and $y_2:x\rightarrow x$. Assume now that $x\in\mathbb{R}^+$ then for all $x\in(0,\infty)$, $y_1$ is a continuous bijective mapping. Similarly it is true for $y_2$. Then clearly $y=y_1y_2\in(0,\infty)$.
Now assume/given that $x\in(-\infty,0)$, then cleary $y_1$ is a continuous bijective mapping from $(-\infty,0)$ to $(0,\infty)$ and $y_2$ is a continuous bijective mapping from $(-\infty,0)$ to $(-\infty,0)$. Since $y_1$ and $y_2$ are both continuous, $y$ is also continuous, and for every $x\in(-\infty,0)$, $y=y_1y_2$ is $\in(-\infty,0)$
As a result of above for all $x\in\mathbb{R}$ we have the conclusion that $x^3\in\mathbb{R}$.
A: Every polynomial of odd degree and with real coefficients must have a real root. This is because every polynomial with real coefficients has complex roots that are conjugate, so the number of non real roots is even hence the number of real roots must  be odd (the Fundamental theorem of algebra states that every polynomial with complex coefficients and of degree $n$ must have exactly $n$ complex roots with possible repetition). Now the polynomial $P=y^3-x$ is of degree three in $y$ (here $x$ is seen as the constant $P(0)$ of the polynomial),  hence $P$ has at least a real root $\alpha$ which means there exists $\alpha\in\mathbb R$ such that $P(\alpha)=\alpha^3-x=0$ hence $\alpha^3=x$. 
A: I think this works too, a bit more mechanical but only makes use of the Intermediate Value Theorem.
Proof
Case 1: If x=0, then y=0.
Case 2: If x > 0, choose f(x) = x^3, a continuous real function in $(-\infty, \infty)$. Let the interval be [0, x+1]. 
Then $f(x+1)=(x+1)^3 = x^3+3x^2+3x+1>x>f(0)=0.$
From the Intermediate Value Theorem we know there exists a real number y in the interval [0, x+1] such that $f(y)+y^3=x$.
Case 3: If x<0, choose f(x) = x^3, a continuous real function in the interval $[x-1, 0]$. Then $f(x-1) = x^3-3x^2+3x-1$ and it follows that $f(x-1) = x^3-3x^2+3x-1<x<0=f(0)$.
Again from the Intermediate Value theorem, for each $x<0$ there exists a real number y in $[x-1, 0]$ such that $f(y) = y^3=x$.
The three cases exhaust all possible values of x.
