⑴化简：[(x+1/x)ˆ6-(xˆ6=1/xˆ6)-2]/[(x+1/x)³+(x³+1/x³](请按分式的解法来解）⑵若a²-3a+1=0,则a³/(aˆ6+1)=⑶若x+y+z≠0,且x/(y+z)=y/(z+x)=z/(x+y),则x/(x+y+z)=

⑴化简：[(x+1/x)ˆ6-(xˆ6=1/xˆ6)-2]/[(x+1/x)³+(x³+1/x³](请按分式的解法来解）⑵若a²-3a+1=0,则a³/(aˆ6+1)=⑶若x+y+z≠0,且x/(y+z)=y/(z+x)=z/(x+y),则x/(x+y+z)=
⑴化简：[(x+1/x)ˆ6-(xˆ6=1/xˆ6)-2]/[(x+1/x)³+(x³+1/x³](请按分式的解法来解）
⑵若a²-3a+1=0,则a³/(aˆ6+1)=
⑶若x+y+z≠0,且x/(y+z)=y/(z+x)=z/(x+y),则x/(x+y+z)=

⑴化简：[(x+1/x)ˆ6-(xˆ6=1/xˆ6)-2]/[(x+1/x)³+(x³+1/x³](请按分式的解法来解）⑵若a²-3a+1=0,则a³/(aˆ6+1)=⑶若x+y+z≠0,且x/(y+z)=y/(z+x)=z/(x+y),则x/(x+y+z)=
第三题：
不妨设x/(y+z)=y/(z+x)=z/(x+y)=a
则x=a（y+z)
y=a(x+z)
z=a(x+y)
所以易得原式=（y+z)/2(x+y+z)