xy=35,yz=41,zx=15,solve xyz?

xy = 35 ...... [1]
yz = 41 ...... [2]
zx = 15 ...... [3]

[1]*[2]*[3] :
(xyz)² = 35 × 41 × 15
xyz = ±√(35 × 41 × 15)
xyz = ±5√861

Thus, xyz = 5√861 or xyz = -5√861

xy = 35

y = 35/x


yz = 41

z = 41/y → recall: y = 35/x

z = 41/(35/x)

z = 41x/35


zx = 15

x = 15/z → recall: z = 41x/35

x = 15/(41x/35)

x = 525/(41x)

x² = 525/41

x² = 25 * (21/41)

x = ± 5√(21/41)



First case: x = 5√(21/41)

Recall: y = 35/x = 35/[5√(21/41)] = 7/√(21/41)

Recall: z = 41x/35 = [41 * 5√(21/41)]/35 = 41.√(21/41)/7

= xyz

= [5√(21/41)] * [7/√(21/41)] * [41.√(21/41)/7]

= [5] * [7] * [41.√(21/41)/7]

= 205.√(21/41)


Second case: x = - 5√(21/41)

Recall: y = 35/x = 35/- [5√(21/41)] = - 7/√(21/41)

Recall: z = 41x/35 = [41 * - 5√(21/41)]/35 = - 41.√(21/41)/7

= xyz

= [- 5√(21/41)] * [- 7/√(21/41)] * [- 41.√(21/41)/7]

= - [5] * [7] * [41.√(21/41)/7]

= - 205.√(21/41)


→ xyz = ± 205.√(21/41)

x=35/y y=41/z z=15/x