Why isn't $y=(x^6)^{1/3}$ a polynomial function? I've been told that $y=(x^6)^{1/3}$ isn't a polynomial function because of the radical but I believe that the equation could be simplified to $y=x^2$ which fits the definition of a polynomial function.
 A: Raising to non integer exponents poses some problems. One might define $x^{1/3}$ also for negative values of $x$, because this is essentially the cubic root; but while $\sqrt[3]{x}$ is non ambiguous, a fraction doesn't change according to its representation:
$$
\frac{1}{3}=\frac{2}{6},
$$
but writing $x^{1/3}=x^{2/6}$ at least raises some doubts.
My personal opinion is that powers with non integer exponents should always be bound to positive values of the base, thus making algebraic manipulations possible without restrictions. So, if we bind $x$ to be positive, we have the true identities
$$
x^{1/3}=x^{2/6}=(x^2)^{1/6}=(x^{1/6})^2,
$$
which would be plainly false if $x<0$ were allowed.
I acknowledge that others don't think this way and define $x^{p/q}$ ($p$ and $q$ integers, $q\ne0$) also for negative values of $x$ provided $p$ and $q$ are coprime and $q$ is odd.
What's the convention used in a textbook should be clearly expressed from the beginning. There is no law cast in the stone about this (and most of mathematics, either). Different fields of math use different conventions; just think to the concept of function itself: in several fields functions must have a well defined domain and codomain, while in Analysis this is not strictly enforced. It's not a problem, provided one is made aware of the convention used.
In your case there is no doubt: the function $x\mapsto(x^6)^{1/3}$ assumes, for every real $x$, the same value as the function $x\mapsto x^2$. Thus the two functions are equal and, being $x\mapsto x^2$ clearly a polynomial function, the conclusion is obvious.
On the other hand, $x\mapsto(x^2)^{1/2}$ is not a polynomial function, because it is the same as $x\mapsto|x|$ and there's no polynomial form of this function: polynomial functions are everywhere differentiable, $x\mapsto|x|$ is not differentiable at $0$.
A: sAt least in $\mathbb{C}$ the functions $f(x) := (x^6)^\frac{1}{3}$ and $g(x) := x^2$ are not equal, as $f(\sqrt[3]{i}) = (i^2)^\frac{1}{3} = -1$ and $g(\sqrt[3]{i}) \not\in \mathbb{R}$.
Edit: Oops. As the comments stare $f$ is not well defined in $\mathbb{C}$, so my answer does not work out so easy. Sorry
A: Salman Khan talks about that in one of his (many) videos.
It's at the beginning of this video:
About polynomials
Hope it helps you understanding polynomials.
A: I'm going to be a little pedantic (but that's my job). 
I want to distinguish between a function and the way that the function is specified. The functions
$$
f_1 : \mathbb R \to \mathbb R: x \mapsto x \\ 
f_2 : \mathbb R \to \mathbb R: x \mapsto (x+1) - 1
$$
are identical as functions (they have the same domain, codomain, and relation), but they've been specified differently.  
Most often, we see functions specified by relatively simple expressions, i.e., strings of characters that we recognize as syntactically correct and semantically valid stuff. But sometimes they're specified by tables, or by "cases" (one expression gets used for $x > 2$, another for $x \le 2$, for instance), etc. 
Certain expressions are called "polynomials in $x$"; they consist of a finite sum of terms, each term being either (a) a constant, or (b) the product of a constant (which may be omitted if it's $1$) and a positive integer power of $x$. Terms where the constant is negative can have an "addition" replaced by a subtraction, so that $4 + -2x^2$ can be written $4 - 2x^2$. 
The expression $(x^6)^{\frac{1}{3}}$ doesn't meet the criteria for being a polynomial. The expression $x^2$ does meet the criteria. If I had been the one talking to you about this thing (which would have meant that I was in a very formal mood), I would have said "The function description $y = (x^6)^\frac{1}{3}$ does not define the function via a polynomial expression. It is, however, identical, as a function, to one that can be defined via a polynomial expression." 
In practice, this almost never matters. 
We tend to be sloppy and say that the function "isn't polynomial" when we mean that the description's not a polynomial, and everyone kinda gets it, and we move on. In the same way, we sometimes say "The vector $v$ is a linear combination of the vectors $x_1$ and $x_2$." What we really mean is that there's a linear-combination-expression whose value happens to be $v$, but the distinction almost never matters in much of mathematics. I'm guessing that in logic and formal semantics, it matters quite a lot, but those aren't my areas, so I cannot say.  
