Product of a Triple

$abc=6^4$

$abc=2^4\times3^4$

Let $a=2^{a_1}\times3^{a_2}$ , $b=2^{b_1}\times3^{b_2}$ , $c=2^{c_1}\times3^{c_2}$

Then the equation becomes

$(2^{a_1}\times3^{a_2})(2^{b_1}\times3^{b_2})(2^{c_1}\times3^{c_2})=2^4\times3^4$

$2^{a_1+b_1+c_1}\times3^{a_2+b_2+c_2}=2^4\times3^4$

Then we have

1. $a_1+b_1+c_1=4$

2. $a_2+b_2+c_2=4$

By the technique Stars and Bars ,

The combinations of $(a_1,b_1,c_1)=(a_2,b_2,c_2)={6\choose2}=15$

Then, we multiply the arrangements of them to find the total combinations of $a_1,a_2,b_1,b_2,c_1,c_2$ , which is

$15^2=225$