x^3 > x^2 only if x^3 - x^2 >0.
Then you can factor in x^2 to get that the initial inequality stands only if: x^2(x - 1) > 0. Since x^2 is always positive, then x - 1 must also be positive in order to respect the inequality. This means that what you need to do is establish your x's position from 1: is it greater than 1 or is it not?
1. tells you that x is positive, but that's not enough. You need to know whether it's also greater than 1.
2. x^2 > x means that x^2 - x > 0 or that x(x - 1) > 0, which in turn tells us that x and x - 1 share the sign. They're either both positive or both negative:
a. if they're both negative:
x < 0
x - 1 < 0 --- x < 1
In this case, then indeed we have established that x is smaller than 1.
b. if they're both positive:
x > 0
x - 1 > 0
In this case, x will be greater than 1.
Since we have two cases for stmt 2, then it's not enough either.
Put the two together to eliminate case a from stmt 2 (since x > 0) and get a definitive answer: yes, x is greater than 1.
Inequalities (I)
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maihuna
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One needs to confirm following two:skim333 wrote:Is x^3 > x^2?
(1) x > 0
(2) x^2 > x
OA is C
x is positive as otherwise cube<square for a given number
and x is not in range 0<x<1 as otherwise what we will have is x^3 will be much smaller, for ex: x=1/2, x^3 = 1/8, x^2 = 1/4
x^2 >x => x(x-1)>0 =>x>1 or x<-1
x>0 combined with above says x>1 which is what we r looking here.h
Charged up again to beat the beast 
