∵tanx=2 ==>sinx=2cosx.(1)
==>(2cosx)²+cosx=1
==>cos²x=1/5.(2)
∴sinx/(sin³x-cos³x)=sinx/[(sinx-cosx)(sin²x+sinxcosx+cos²x)]
=2cosx/[cosx(1+2cos²x)] (由(1)代换)
=2/(1+2cos²x)
=2/(1+2*(1/5)) (由(2)代换)
=10/7.