cornell2005 wrote:I get C also, but am not sure how to prove that 1 and 2 together are sufficient. Anyone have any ideas? Thanks.

the presence of x^(1/2), also known as √x, in the problem guarantees that x is nonnegative. therefore, there are four cases:
(1) x = 0
(2) 0 < x < 1
(3) x = 1
(4) x > 1

cases (1) and (3): in these cases, &#8730;x = x^3 = x. therefore, saying that any one of these three quantities is greater than y is equivalent to saying that any other one of them is equivalent to y.

**case (2): if 0 < x < 1, then x^3 < x < &#8730;x. if both statements are true, then y must be smaller than both x^3 and &#8730;x, so that we have y < x^3 < x < &#8730;x.

***case (4): if x > 1, then &#8730;x < x < x^3 (the "normal" order of these three quantities). if both statements are true, then y must be smaller than both x^3 and &#8730;x, so that we have y < &#8730;x < x < x^3.

in all four cases, statements 1 and 2 together are sufficient, so answer = c.

--

**note: this is the case that proves statement 1 insufficient, as y could be less than &#8730;x but more than x.

***note: this is the case that proves statement 2 insufficient, as y could be less than x^3 but more than x.