Is it possible to change $x^3\prod_{n=1}^{\infty}\left(1-\frac{x^4} {2\pi^2n^2}-\frac{x^6}{2\pi^3n^3}\right)$ to equal $x\sin{(x^2)}$? Consider a function $$f(x)=x^3\prod_{n=1}^{\infty}\left(1-\frac{x^4} {2\pi^2n^2}-\frac{x^6}{2\pi^3n^3}\right)$$
This function has its zeros when $x=0$ or $x^2=k\pi$.
This function is very similar to another function which is $$g(x)=x\sin{(x^2)}$$. They have the same zeros and just that their increasing speed is different.
Is it possible to make some changes such that they are equal? One seems to be increasing and one is decreasing. Also, they are the functions that appears to have the exponential function $\frac 1 {x^3}$.

Note :
Consider the Weierstass factorization of $\sin z$ :
$$\sin z = z \cdot \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{\pi^2 n^2}\right).$$
let $z = x^2$ then
$$\sin(x^2 ) = x^2 \cdot \prod_{n=1}^{\infty} \left( 1 - \frac{x^4}{\pi^2 n^2} \right)$$
and
$$x \sin(x^2 ) = x^3 \cdot \prod_{n=1}^{\infty} \left( 1 - \frac{x^4}{\pi^2 n^2} \right).$$
