Is x greater than x^3 ?

(1) x is negative

(2) x^2 - x^3 > 2

Is x greater than \(x^3\) ?

(1) x is negative.

(2) \(x^2- x^3 > 2\)

Bunuel, can you explain Stat 2...I did half way ..\(x^2\) ( 1 - x ) > 2 and taken \(x^2\) > 0 as it is square root and tried with 1-x >2...and struck here.

This is a really old question, and I found it recently in Jeff Sackmann's Data Sufficiency Challenge question bank. I am afraid I don't understand why Statement (2) above is sufficient, particularly the highlighted part above.

My interpretation is this: In order to keep x^2 > x^3, (1-x) > 0 (i.e. positive). This implies that 1 > x. x could therefore be negative integer, negative fraction or positive fraction. For instance:

x = -2 => x^2 = 4 and x^3 = -8 => x^2 > x^3 and 1-(-2) > 0. So, x > x^3

x = -1/2 => x^2 = 1/4 and x^3 = -1/8 => x^2 > x^3 and 1-(-1/2) > 0. However, x < x^3

x = 1/2 => x^2 = 1/4 and x^3 = 1/8 => x^2 > x^3 and 1-(1/2) > 0. So, x > x^3

Clearly, x CAN be fractional values between -1 and 1 for x^2 > x^3. But each time, the relation between x and x^3 may change. So B is not sufficient.

Can somebody please explain where I am going wrong? Or provide an alternative explanation. Thanks a lot!

Is x greater than \(x^3\) ?

(1) x is negative.

(2) \(x^2- x^3 > 2\)


I find such questions challenging. Please explain.

DS : x>x^3?

Statement 1 : x<0
-1<x<0 , x^3 > x
x<-1, x<x^3

NOT SUFFICIENT


Statement 2 : x^2 - x^3 > 2
x^2> 2+x^3
For x>1, x^3 > x^2 So, x^3 +2 > x^2. This is not the case
For 0<x<1, x^2 <1 So, x^2 <2 +x^3 This is also not the case.
For -1<x<0, x^2<1 , -1<x^3<0 So, x^2 <2 +x^3 This is also not the case.
For x = -1 , x^2=2+x^3 This is also not the case.
For x<-1, x^2>2+x^3 This is the case.

Now if x<-1, Clearly x>x^3

Hence Answer B
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