已知ab不等于0，a与b互为相反数，s与t互为倒数，求a^3+b^3/a^3-b^3+s+t/st^2+s^2t的值

是已知ab不等于0，a与b互为相反数，s与t互为倒数，求（a^3+b^3）/（a^3-b^3）+（s+t）/（st^2+s^2t）的值
吗？a+b=0;st=1
=[(a+b)（a*a-ab+b*b)]/([(a-b)(a*a+ab+b*b)+(s+t)/[st(s+t)]
=0+1/st
=0+1
=1

(a^3+b^2)/(a^3-b^3)+(s+t)/(st^2+s^2t)
=(a+b)(a^2-ab+b^2)/(a-b)(a^2+ab+b^2)+(s+t)/st(s+t)
=0+1
=1

解：a+b=0,st=1.
a^3+b^3/a^3-b^3+s+t/st^2+s^2t
=a^3-b^3+b^3/a^3+s+t/t+ts^2
=0-1+s+1+s
=2s

1