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 :)

Consider $f(x)=2^5(x+\dfrac{2}{2})(x+\dfrac{3}{2})(x+\dfrac{5}{2})(x+\dfrac{7}{2})(x+\dfrac{11}{2})$ Your sum equals $3f(1)=3[2^5(1+\dfrac{2}{2})(1+\dfrac{3}{2})(1+\dfrac{5}{2})(1+\dfrac{7}{2})(1+\dfrac{11}{2})]=3(2+2)(2+3)(2+5)(2+7)(2+11)$
