Since we know GRE takes into account the positive value only for the square root of any number, then option B should be correct as well with that logic.

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I donâ€™t think you need to care about +ve or -ve scenario with this one!

\sqrt[3]{x^9}= x^{9 \times{\frac{1}{3}}} = x^3

So option B is right as well?

(But when substituting a value, I realized Option B doesnâ€™t match. )

Even I had the same exact thought pop up while listening to Gregâ€™s explanation. @gregmat , can you please confirm whether this is correct or not ?

In option B, you can plug in a negative x value to your heartâ€™s content because the even exponent makes it positive before the root is applied.

Notice these two graphs are not the same:

Also, GRE does consider negatives under the radical if the root is odd