x,y,z are positive real numbers such that x+y+z=9 and xy+9xz+25yz=9xyz. Find the sum of all possible values of xyz.

Answer 1

Try all x,y,z combinations that work. There will not be very many.
(x,y,z) 9xyz xy+9xz+25yz
(1,7,1) 63 >63
(1,1,7) 63 >63
(2,1,6) 108 >108
(2,6,1) 108 >108
(2,5,2) 180 >180
(2,2,5) 180 >180
(2,4,3) 216 >216
(2,3,4) 216 >216
(3,3,3) 243 225 + 81 + 9 > 243
(3,4,2) 216 200 + 54 + 12 > 216
(3,2,4) 216 200 + 108 + 6 > 216
(3,1,5) 235 125 + 135 + 9 > 235
(3,5,1) 235 125 + 27 + 15 < 235
(4,3,2) 216 150 + 72 + 12 > 216
(4,2,3) 216 150 + 108 + 8 > 216
(4,1,4) 144 100 + 144 + 4 > 144
(4,4,1) 144 100 + 36 + 16 > 144
(5,3,1) 135 75 + 45 + 15 = 135
(5,1,3) 135 75 + 135 + 6 > 135
(5,2,2) 180 100 + 90 + 10 > 180
(6,1,2) 108 50 + 108 + 6 > 108
(6,2,1) 108 50 + 54 + 12 > 108
(7,1,1) 63 25 + 63 + 7 > 63

The only solution is (x,y,z) = (5,3,1), for which xyz = 15

Answer 2

that is such a mission though lol, is that the only way?

Answer 3

Though the answer may be correct, the solution isn't ,because x,y,z are real numbers not integers