∵0<x<1,分2种情况讨论, ①当3a>1,即a>时, |log3a(1-x)3|-|log3a(1+x)3| =|3log3a(1-x)|-|3log3a(1+x)| =3[-log3a(1-x)-log3a(1+x)] =-3log3a(1-x2). ∵0<1-x2<1,∴-3log3a(1-x2)>0. ②当0<3a<1,即0<a<时, |log3a(1-x)3|-|log3a(1+x)3| =3[log3a(1-x)+log3a(1+x)] =3log3a(1-x2)>0. 综上所述,|log3a(1-x)3|>|log3a(1+x) |