Given: x³ < x²
This tells us that x ≠ 0. So, we can be certain that x² is POSITIVE.
Since x² is POSITIVE, we can safely divide both sides of the inequality by x²
When we do this, we get: (x³)/(x²) < 1
Simplify: x < 1
Subtract 1 from both sides to get: x - 1 < 0
In other words, x - 1 is NEGATIVE

Answer: D

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Bunuel is there a mistake in the question?
Simplifying gives x^2(x-1)<0 so x<1 and x!=0 so there are no correct answers... also, options (D), (E) are identical


EDIT, to answer corrected question:
As all we're given are equations, we'll work with them using simplification tools.
This is a Precise approach.

Simplifying our stem gives x^3 - x^2 < 0 so x^2(x - 1) < 0. Since x^2 is always non-negative then x - 1 < 0 and x < 1. Also, x cannot be 0, otherwise we would get 0 < 0.
A. could be 1/2 so it is false.
B. could be -(-1)=1 and is false.
C. could be (1/2)^5 > 0 and is false
D. as x < 1 then x - 1 < 0. Yes!

(D) is our answer.
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Edited. Thank you.
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Alternatively, we can TEST some values.

We're told that x^3 < x^2
So, one possible value is x = -1

Now let's test the answer choices. We get:
A. -1, which is negative. KEEP
B. −(-1) =1, which is positive. ELIMINATE
C. (-1)^5 = -1, which is negative. KEEP
D. (-1) − 1 = -2, which is negative. KEEP
E. (-1)^(−1) = 1/(-1) = -1, which is negative. KEEP

Okay, we're still left with A, C, D and E

We need to test another value.
Another possible value is x = 1/2
Test the remaining answer choices...
A. 1/2, which is positive. ELIMINATE
C. (1/2)^5 = 1/32, which is positive. ELIMINATE
D. (1/2) − 1 = -1/2, which is negative. KEEP
E. (1/2)^(−1) = 2/1 = 2, which is positive. ELIMINATE

By the process of elimination, the correct answer is D

Cheers,
Brent
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For x^3 < X^2 to be true, there are 3 cases
Case 1: x< -1, in which case except B all are negative
Case 2: -1<x<0, in which case except B, all are negative
Case 3: 0<x<1, in which case except A,C,E all are negative.

Thus in all the 3 case above for x-1 is negative for which is D is negative.

hence the answer is D

X^3-x^2<0
x^2(x-1)<0

X^2 is always positive
X-1 is negative

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Since x^3 is less than x^2, there are two cases for x: 1) x could be a negative number, or 2) x could be a (positive) number whose value is between 0 and 1.

We see that, in either case, x - 1 will always be negative.

Answer: D
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Good way is to test standard values 0 ,2,1/2 ,-1/2 ,-2. These values always help in these kind of questions , so that we don't miss any case
as now , x^3 <x^2 , it means x can be 1/2 ,-1/2 ,-2
Option
A : X can be 1/2 : incorrect
B. −x : can be -1/2 or -2 : will give positive result: incorrect
C. x^5 : in case of 1/2 positive: incorrect
D. x − 1 : negative in All cases: One can check all solutions
E. x^(−1) : positive in case of 1/2 : Incorrect

Solution

Here, it is given that \(X^3\) < \(X^2\)
\(X^3\)-\(X^2\) < 0
\(X^2\) (X-1) < 0

No matter the value of X (either + or -), \(X^2\) will always be Positive.
So, for \(X^2\)(X-1) < 0, (X-1) should be negative.

Answer D

(X)^3 < (X)^2

there are 2 Cases in which (X)^3 will be LESS THAN < (X)^2


Case 1: when X is a (+)Pos. Proper Fraction from 0 < X < 1 ------> Raising a (+)Pos. Proper Fraction to a HIGHER Integer Exponent will result in a DECREASE in Value

or

Case 2: when X is ANY (-)Negative Value EXCEPT (-)1


-A- X

does NOT have to be (-)Neg. if X = 1/2


-B- (-)X

does NOT have to be (-)Neg. if X = -5

(-)-5 = +5


-C- X^5

if X = +1/2 -----> does NOT have to be (-)Negative


-E- (X)^ -1

Rule of (-)Negative Integer Exponents: you take the Reciprocal of the Base and change the Sign of the (-)Neg. Integer Exponent

(X)^ -1 = (1/X)^1 = 1/X

if X = +1/2 -----> 1/X = 2, which is NOT Negative


-D- X - 1

case 1: if X is a +Pos. Proper Fraction, the result will be (-)Negative

and

case 2: any (-)Negative Value combined with another (-)Negative Value will result in a (-)Negative Value



-D- is the correct answer