If \(x≠0\), is \(-\frac{4}{x^3}\) negative?

Notice that the question basically asks whether x is a positive number, because if it is then \(-\frac{4}{x^3} =-(\frac{positive}{positive})=negative\).

(1) \(x\) is positive. Directly answers the question. Sufficient.

(2) \(x^3 - x^2\) is positive --> x^3- x^2 > 0 --> x^3 > x^2 --> reduce by x^2 (we can safely do that because x^2 will be positive): x > 1. The same here: x is positive. Sufficient.

Answer: D.

Hope it's clear.
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Hi Bunuel,

can we safely say that if Cube of a number - square of a number is positive, then that number is positive.
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Hi Bunuel,
I have a slight query. Could it not be that x^3 - x^2 > 0 instead of > 1.
Even if say x is 1/2, the result would be a negative number. But that negative sign cancels out with the negative sign mentioned in the equation and the result is positive anyway?

I just wanted to make sure so I don't mess up while assuming values for such problems.
Please help!
Thanks


If \(x≠0\), is \(-\frac{4}{x^3}\) negative?

Notice that the question basically asks whether x is a positive number, because if it is then \(-\frac{4}{x^3} =-(\frac{positive}{positive})=negative\).

(1) \(x\) is positive. Directly answers the question. Sufficient.

(2) \(x^3 - x^2\) is positive --> x^3- x^2 > 0 --> x^3 > x^2 --> reduce by x^2 (we can safely do that because x^2 will be positive): x > 1. The same here: x is positive. Sufficient.

Answer: D.