Stuck at evaluating $\sum\limits_{d \mid n} \tau(d)$ It seems the number of nonnegative integer solutions to the equation $xyz=n$ is given by 
$$\sum\limits_{d \mid n} \tau(d)$$
$\tau$ is the number of divisors function. I'm wondering if there is a way to simplify this sum. Really appreciate any kind of help. Thank you.

Here is my attempt so far
$$xyz = n$$
$x$ can be any of the factors of $n$ and the product $yz$ will be $n/x$.
Since $yz$ sees all the factors of $n$, the number of nonnegative integer solutions to $xyz=n$ is simply the sum of divisors of the product $yz$.
Edit : Special thanks to @Tryss for identifying an error in the formula. I've fixed it now..

Factor $n$ as $p_1^{\alpha_1}\cdot\ldots\cdot p_k^{\alpha_k}$. Then every solution is associated with three vectors (the exponents in the factorizations of $x,y,z$) with non-negative integer components and sum given by $(\alpha_1,\ldots,\alpha_k)$. By stars and bars, it follows that the number of solutions is given by 
$$ \prod_{h=1}^{k}\frac{(\alpha_h+2)(\alpha_h+1)}{2}.$$
