Determining properties of a polynomial $f$ satisfying $f(x^2)-xf(x) = x^4(x^2-1)$ for $x \in\Bbb R^+$ 
Let $f$ be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x \in\Bbb R^+$. Then which of the following is correct?
A) $f$ is an even function
B) $f$ is an odd function
C) $\displaystyle\lim_{x\to \infty} \frac{f(x)}{x^3}=1$
D) $\displaystyle\lim_{x\to \infty} \left(\frac{f(x)}{x^2}-x \right)$ exist and is equal to a non-zero quantity.

I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C). 
Any help would be appreciated.
 A: Some questions have been raised in the comments about my answer below. A polynomial  on $\mathbb  R$ which satisfies the given functional equation on $[0,\infty)$ necessarily has the form $ax+x^{3}$ on that interval. But then it must have the same form on  all of $\mathbb  R$. So I think the answer below is correct. 
Look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
A: You don't need to find $f(x)$. For instance, statements A and B can be dealt with by considering
$$
f(x)=\frac{f(x^2)-x^4(x^2-1)}{x} \tag{*}
$$
and so
$$
f(-x)=\frac{f(x^2)-x^4(x^2-1)}{-x}=-f(x)
$$
showing that $f$ is an odd function. One can object that (*) only holds for $x>0$, but the right-hand side is a polynomial, (implying $f(0)=0$) and if two polynomials agree on an infinite set they're equal.
If the degree of $f$ is $n$, then the degree of $f(x^2)$ is $2n$. By (*), the degree of $f(x^2)$ must be $6$, or the equality could not hold. Hence $n=3$.
The function $f(x)/x^3$ has finite limit $l$, owing to $\deg f(x)=3$. Now
$$
\frac{f(x^2)-xf(x)}{x^6}=\frac{f(x^2)}{(x^2)^3}-\frac{1}{x^2}\frac{f(x)}{x^3}=\frac{x^2-1}{x^2}=1-\frac{1}{x^2}
$$
Hence $l=1$.
Similarly,
$$
\frac{f(x)}{x^2}-x=\frac{f(x^2)}{x^3}-x(x^2-1)-x=\frac{f(x^2)}{x^3}-x^3=
x\left(\frac{f(x^2)}{(x^2)^2}-x^2\right)
$$
Now it's clear that statement D is false: the limit exists (finite or infinite), but if it's finite it must be $0$.
Therefore B and C are true. This allows us to find $f(x)$: it is a degree $3$ polynomial, with leading coefficient $1$ and no term of even degree. Hence we have
$$
f(x)=x^3+ax
$$
Apply the functional equation:
$$
f(x^2)-xf(x)=x^6+ax^2-x^4-ax^2=x^4(x^2-1)
$$
holds for every $x$. Therefore $a$ can be anything.
A: To expand on @KaviRamaMurthy's answer:
Let $f=\sum^n a_kx^k$ where $a_n\ne 0$ for $x\ge 0$. Then
$$\sum a_kx^{2k}-\sum a_kx^{k+1}=x^6-x^4$$
$$\sum a_kx^{2k}=\sum a_kx^{k+1}+x^6-x^4$$
So $2n\in [0,n+1]\cup\{6\}$ which means $n\le 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $x\ge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant

for $x\in R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.

