inequality

[bose]

a^3>b^3 then a> b but if a^-3 > b^-3 i do not think we can conclude a<b since we cannot take reciprocal.ex-a=2 b=4 a<b and a=4, b=-2 a> b
makumajon can u pls confirm that i am correct?

[amitabh007]

a^3-b^3>0
(a-b)(a^2+b^2+ab)>0......
of the two product parts...latter is always +ve
hence the inequality becomes...a-b>0
a>b..

now for...a^-3-b^-3>0...
(b^3-a^3)/(a^3*b^3)>0...multiply ab to both numerator and denominator
....ab(b^3-a^3)>0........
..ab(b-a)(b^2+a^2+ab)>0...here the third part in the product is always +ve
hence inequality becomes...ab(b-a)>o

take case1...a>0
eq becomes....b(b-a)>0......means either b<0 or b>a...for a>0
which you have proved for a=2,b=4 and for a=2,b=4

take case2...a<0
eq becomes...b(b-a)<0....means a<b<0
for ex...a=-1,b=-1/2

hence...i think u r correct.

[shyamprasadrao]

nothin can be concluded from the statement... a^-3 > b^-3 ... probably if a mod is added to them.. you can state b>a

[shyamprasadrao]

and if they are integersss