It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$? 
It is given that $x.y.z=12^3$ where $x,y,z$ are positive integers.Then , how many different solutions are there for $x+y+z$?

Its answer is easy if it were $x.y=12^3$ , because there are $28$ possible $(a,b)$ pairs by $x.y=2^6 \times 3^3$ and $C(8,2) \times C(4,1) =28$.
Moreover, half of $28$ will give us the number of $x+y$ thanks to symmetry property.
Let's come to original question:
Firtly , i found the number of how many possible $x,y,x$ there exist by combination with repetition such that $C(6+3-1,6) \times C(3+3-1,3)=280$.
However , when i want to find the number of different values of $x+y+z$ , i stuck in.Because , i thought that if i divide $280$ by $3$ , i can obtain the result.However , it did not work and i dont know what i should do.
Therefore, i hope to find tricks or solutions for my problem. Moreover , what can i do for the expanding versions of this question such as $x+y+z+t$

Let's define an order that $a < b$ if the power of $3$ that divides $a$ is less than the power of $3$ that divide $a$ or those powers are equal then if power of $2$ that divides $a$ is larger then the power than divides $b$
Then wolog $x \le y \le z$ and we need to consider the cases
$x=y=z$
$x=y < z$
$x < y = z$
$x < y < z$.
Case 1: $x = y = z$ so $3^1|x,y,z$ and $2^2|x,y,z$ and $x=y=z=12$ and $x+y+z = 36$.
Case 2: $x=y < z$ then either $3|x,y,z$ and $2^3|x,y$ or $3^3|z$ and $1,2,4|x,y$
That gives us $4$ more options $x=y=24;z=3$ or $x=y=1,2,4; z = 27\cdot(2^6,2^3,1)$
So that's $5$ options so far.
Case 3: $x< y = z$ then $3|x,y,z$ and either $2^6|x$ or $2^4|x, 2|y,z$
So that's $2$ more options. $7$ so far.
Case 4: $x < y < z$.
either $3|y; 3^2|z$ or $3|x,y,z$ and $2^4|x; 2^2|y$ or $2^3|x; 2^2|y; 2|z$
The latter two give us $2$ more for $9$.
And finally The first case we can have the powers of $2$ distributed any way.
That $3$ times $1,1,2^6$.
$1$ times $1,2^2,2^4$ and $2$ times $1,2^3,2^3$.
And $2$ times $2,2,2^4$.  And $6$ times $2,2^2,2^3$.
And $1$ times $2^2,2^2,2^2$ so that is $3+1+2+2+6+1=15$.
So a total of $24$.
I'm taking it on faith, perhaps incorrectly,  that we can show these solutions are pairwise relatively prime so distinct.
And as Robert Isreal got $49$ sums I'm obviously making an error. Probably missing some options.
Still I'll leave this up as a strategy that I think is valid.
