Is it possible to open up this particular convolution equation? I have three vectors, $x[n]$, $w[n]$, and $z[n]$. I need to compute:

$y[n] = z[n] * \Big(x[n] \cdot w[n] \Big)$. 

This is easy to do. However, is there a way I can do this by 'opening' up the bracket? 

For example, let us say that I am already given $z[n] * x[n]$. If this was the starting point, what could I do with this and with the three already given vectors to get $y[n]$? The wiki on convolution properties does not seem to have a case for a scenario like this.
Here is a re-write of a previously mistaken answer:

Let us write $y[n] = z[n] * \Big(x[n] \cdot w[n] \Big) = \displaystyle\sum_{m=-M}^{M} {z[n-m] \cdot (x[m] \cdot w[m])}$ and rearrange as $y[n] = \displaystyle\sum_{m=-M}^{M} {(z[n-m] \cdot x[m]) \cdot (w[m])}$. After this point, we cannot take the terms out of the brackets.