<s>[INST] If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then Find $6xyz$. If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then
Find $6xyz$.  
How can I approach this problem?
I need some hints. Thanks. 
 [/INST] $x + y + xy = 3  \implies    ( 1 + x ) ( 1 + y ) = 4\tag1$
$y + z + yz = 8  \implies   ( 1 + y ) ( 1 + z ) = 9\tag2$
$z + x + zx = 15  \implies   ( 1 + z ) ( 1 + x ) = 16\tag3$
Multiplying $(1)$ and $(3)$, we have
$$( 1 + x )^2 ( 1 + y )( 1 + z )=4\cdot 16\implies ( 1 + x )^2= \dfrac{64}{9}\qquad \text{{using $(2)$}}.$$
$$\implies x=\pm \dfrac{8}{3}-1=\dfrac{5}{3}~, -\dfrac{11}{3}$$
From $(1)$, $~y=-1+\dfrac{4}{1+x}=\dfrac{1}{2},~-\dfrac{5}{2}~$ respectively for $~x=\dfrac{5}{3}~, -\dfrac{11}{3}~$
From $(3)$, $~z=-1+\dfrac{16}{1+x}=5,~-7~$ respectively for $~x=\dfrac{5}{3}~, -\dfrac{11}{3}~$
So case I:  When $~x=\dfrac{5}{3},~y=\dfrac{1}{2},~z=5~,$
$$~6xyz=6\times \dfrac{5}{3} \times\dfrac{1}{2}\times 5=25~.$$
and case II: When $~x=-\dfrac{11}{3},~y=-\dfrac{5}{2},~z=-7~,$
$$~6xyz=6\times (-)\dfrac{11}{3} \times(-)\dfrac{5}{2}\times(-) 7=-385~.$$ 
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