<s>[INST] On the Sum of Some of the Factors of $2310$ 
Given 3 natural numbers  a,b,c  such that $a\times b\times c=2310$. Find the sum of $\sum a+b+c$

My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
So now since $a,b,c$ are symmetric , their sum should be the same , and I should get , $\sum a+b+c= 3 \sum a$
But computing this is becoming tedious. As when $a$ is $1$ ,then $b,c$ will have $2^5 $combinations. When $a = 2$ or $3$ or $5$ or $7$ or $11$ ,then  $b,c$ will have $2^4$ combination. So, the sum. Will become $$3 \sum a =  3 \{2^5 + (2+3+5+7+11) 2 ^4 + (2×3 +2×5... +3×5 +....) 2^3 +  (2×3×5 +... ) 2^2 ..... +( 2×3×5×7×11)2^0 \} $$ 
What next ?? , Because this sum really starts to become too big. I'm sure there must be a shorter way as it is an exam type problem supposed to be solved in a limited time. Or is there any better approach ?? Please help :)
 [/INST] Silly overkill method:
For any (positive) natural number $N$, set
$$d_N=\sum_{abc=N}(a+b+c)\text{.}$$
Consider the Dirichlet series
$$f(s)=\sum_{N}\frac{d_N}{N^s}\text{.}$$
Then
$$\begin{split}f(s)&=\sum_{a,b,c}\frac{a+b+c}{a^sb^sc^s}\\
&=3\zeta(s-1)\zeta(s)^2\end{split}$$
so that $f(s)$ admits the Euler product
$$\begin{align}
\frac{f(s)}{3}&=\prod_{p\text{ prime}}f_p(p^{-s}) \\
f_p(T)&=\frac{1}{(1-pT)(1-T)^2}=1+(p+2)T+\ldots
\end{align}$$
Now suppose that $N$ has no repeated prime factors. Then reading off the coefficient of $N^{-s}$ gives
$$d_N=3\prod_{p|N}(p+2)\text{.}$$
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