Switching numerator and denominator

The three equations are equivalent to

$ab+ac=5$

$bc+ab=10$

$ac+bc=13$

Therefore, $2=5+10-13=(ab+ac)+(bc+ab)-(bc+ac)=2ab \implies ab=1$

Similarly, we can get $bc=9,ac=4$ .

Thus, $ab.bc.ac=(abc)^2=36 \implies \boxed{abc=6}$

$5=ab+ac$ $10=ab+bc$ $13=ac+bc$ Adding all three equations yields $28=2(ab+bc+ac)$ Now, we can plug in each of the first equations into this newly derived equation to find that $bc=9, ac=4, ab=1$ .

Therefore, $a=\dfrac{2}{3}, b=\dfrac{3}{2}, c=6$ ,

which multiply to $\left(\dfrac{2}{3}\right) \left(\dfrac{3}{2}\right)\left(6\right)=6$ .