已知实数a,b,c满足方程组,则abc的值是

1/ab+1/ca=-4-1/a^2; ==>1/a^2+1/ab+1/ca=-4
1/bc+1/ab=8-1/b^2; ==>1/ab+1/b^2+1/bc=8
1/ca+1/bc=12-1/c^2; ==>1/ca+1/bc+1/c^2=12
得 1/a(1/a+1/b+1/c) = -4;
1/b(1/a+1/b+1/c) = 8;
1/c(1/a+1/b+1/c) = 12;
再得：
(1/a+1/b+1/c)=-4a=8b=12c;
则：a = -2b = -3c;
1/(-2b)+1/b+1/(2b/3)=8b
-1/(2b)+1/b+3/(2b)=8b
-1+2+3=16b^2
b^2=1/4
b=±1/2
则 abc =-2b*b*2b/3=-4b^3/3=±1/6;