I'm pretty sure on the real GMAT the question would specify that x was positive in the stem. In any case, presenting essentially the same reasoning as Kevin,

\(\text{If } x > 1 \text{, then } \sqrt{x} < x < x^3 \text{ so if } y \text{ is less than both } \sqrt{x} \text{ and } x^3 \text{, } y \text{ must be less than } x \\

\text{If } 0 < x < 1 \text{, then } x^3 < x < \sqrt{x} \text{ so if } y \text{ is less than both } \sqrt{x} \text{ and } x^3 \text{, } y \text{ must be less than } x \\

\text{If } x = 1 \text{ or } 0 \text{, then } x^3 = x = \sqrt{x} \text{ so if } y \text{ is less than both } \sqrt{x} \text{ and } x^3 \text{, } y \text{ must be less than } x \\\)

So C. Note that while every positive number has two square roots, \(\sqrt{x}\) has only one value: the non-negative square root of x.

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