已知A,B互为相反数,X,Y互为倒数,C的绝对值=3,求1/3C^3(XY)^3-2/3C^2(A+B)^2-4/3C^3(XY)^3

1/3C^3(XY)^3-2/3C^2(A+B)^2-4/3C^3(XY)^3
=1/3C^3-0-4/3C^3
=-C^3
=27或-27

因为c的绝对值=3所以c=正负3因为A,B互为相反数所以a+b=0因为X,Y互为倒数所以xy=1

根据题意得:
A+B=0
XY=1
C=3或-3

(1/3)*C^3*(XY)^3-(2/3)*C^2*(A+B)^2-(4/3)*C^3*(XY)^3
=(1/3)*27*1-0-(4/3)*27*1
=9-0-36
=9-36
=-27

(1/3)*C^3*(XY)^3-(2/3)*C^2*(A+B)^2-(4/3)*C^3*(XY)^3
=(1/3)*(-27)*1-0-(4/3)*(-27)*1
=-9-0+36
=-9+36
=27

根据题意得:
A+B=0
XY=1
C=3或-3

(1/3)*C^3*(XY)^3-(2/3)*C^2*(A+B)^2-(4/3)*C^3*(XY)^3
=(1/3)*27*1-0-(4/3)*27*1
=9-0-36
=9-36
=-27

(1/3)*C^3*(XY)^3-(2/3)*C^2*(A+B)^2-(4/3)*C^3*(XY)^3
=(1/3)*(-27)*1-0-(4/3)*(-27)*1
=-9-0+36
=-9+36
=27 就是这样