318.2
จากการย้ายข้างไปมาได้
$\frac{2}{{yz}} + \frac{1}{{y^2 }} + \frac{1}{{z^2 }} = 3 + \frac{1}{x} + \frac{1}{{x^2 }}$
$\frac{2}{{zx}} + \frac{1}{{z^2 }} + \frac{1}{{x^2 }} = 4 + \frac{1}{y} + \frac{1}{{y^2 }}$
$\frac{2}{{xy}} + \frac{1}{{x^2 }} + \frac{1}{{y^2 }} = 3 + \frac{1}{z} + \frac{1}{{z^2 }}$
ให้ $\frac{1}{x} = a,\frac{1}{y} = b,\frac{1}{z} = c$
ได้ $2bc + b^2 + c^2 = 3 + a + a^2 $...(1)
$2ca + c^2 + a^2 = 4 + b + b^2 $...(2)
$2ab + a^2 + b^2 = 3 + c + c^2 $...(3)
นำ (1)+(2)+(3) ได้ $2a^2 + 2b^2 + 2c^2 + 2ab + 2bc + 2ca = 12 + a + b + c + a^2 + b^2 + c^2 $
หรือ $(a + b + c)^2 - (a + b + c) - 12 = 0$
ได้ $a + b + c = 4, - 3$
กรณีัีี 1 $a + b + c = 4$
$(b + c)^2 = 3 + a + a^2 $
$(4 - a)^2 = 3 + a + a^2 $
$a^2 - 8a + 16 = 3 + a + a^2 $
$9a = 13$
$a = \frac{{13}}{9}$
$(c + a)^2 = 4 + b + b^2 $
$(4 - b)^2 = 4 + b + b^2 $
$b^2 - 8b + 16 = 4 + b + b^2 $
$9b = 12$
$b = \frac{4}{3}$
$(a + b)^2 = 5 + c + c^2 $
$(4 - c)^2 = 5 + c + c^2 $
$c^2 - 8c + 16 = 5 + c + c^2 $
$9c = 11$
$c = \frac{{11}}{9}$
กรณี 2 $a + b + c = - 3$
$(b + c)^2 = 3 + a + a^2 $
$( - 3 - a)^2 = 3 + a + a^2 $
$a^2 + 6a + 9 = 3 + a + a^2 $
$5a = -6$
$a = -\frac{6}{5}$
$(c + a)^2 = 4 + b + b^2 $
$( - 3 - b)^2 = 4 + b + b^2 $
$b^2 + 6b + 9 = 4 + b + b^2 $
$5b = -5$
$b = -1$
$(a + b)^2 = 5 + c + c^2 $
$( - 3 - c)^2 = 5 + c + c^2 $
$c^2 + 6c + 9 = 5 + c + c^2 $
$5c = -4$
$c = -\frac{4}{5}$
จาก 2 กรณีได้ว่า
$(a,b,c) = (\frac{{13}}{9},\frac{4}{3},\frac{{11}}{9}),(-\frac{6}{5},-1,-\frac{4}{5})$
ดังนั้น $(x,y,z) = (\frac{9}{{13}},\frac{3}{4},\frac{9}{{11}}),(-\frac{5}{6},-1,-\frac{5}{4})$