A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. Now are
$$\sqrt[3]{y^3},\quad \frac{x^2}{x}\quad\text{or}\quad\left|x\right|\sqrt[2]{x^2}$$
polynomials?
I think $$\sqrt[3]{y^3}\quad\text{and}\quad\left|x\right|\sqrt[2]{x^2}$$ are polynomials (since $x$ may be any number), but $\frac{x^2}{x}$ is not a polynomial (since $x\neq 0$)
On
Well... yes, and no.
The point is: when we say a polynomial, we want so much to entail the fact that it is a finite linear combination of natural powers that it is usual, for example, to define it as a function $f: \mathbb{N} \rightarrow A$ (where $A$ is whatever ring we are on) which is zero for all, except finitely many $a \in A$. (c.f. Lang's Algebra). This is to entail the fact that there exists an "indeterminate", and coefficients. The coefficients are given by $f(n)$ ($f(0)$ is the first coefficient, of what would be $X^0$, $f(1)$ of $X^1$ etc).
If we were going to be nitpicky, even $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x)=x^2$ is not a polynomial. This is what I think @Arthur is trying to entail in his comment. This is what he calls a polynomial function.
However, we of course have no shame on calling this function a polynomial, since there is an obvious equivalence. However, the expressions you mention are not capturing the essence of a polynomial in their form. Personally, I would not refer to them as polynomials, but instead simply as functions. But this is the same thing as refraining from calling a force (on the context of physics) a point on $\mathbb{R}^3$, and instead referring to it as a vector.
In the end of the day, it is all a matter of definitions and communicating oneself.
No, these are not polynomials.