If x,y ,and z are positive real numbers such that x(y+z) = 132, y(z+x) =152 , and z(x+y) = 180, then xyz is Pick one.
A) 40
B) 60
C) 160
D) 100
E) 120
xy+xz = 132 --------------------------------------1)   yz+yx = 152---------------------------------------2)   xz+zy = 180---------------------------------------3)   Re-writing equation 3 as follows:   xz+zy = 152 + 28   xz+zy = yz+yx + 28   xz = yx + 28 ............... (4)   Adding (1)(4)   2xz = 160   xz = 80   xyz has to be multiple of 80,only 160 fits in   Answer = 160
<result>C</result>
If x,y ,and z are positive real numbers such that x(y+z) = 132, y(z+x) =152 , and z(x+y) = 180, then xyz is