First, note that \(a\) is always positive
I. Pick a positive \(a\), assign \( b=0\), and pick a negative \( c\), we get \(a>b>c\). You can choose another case, but overall (I) is possible.
II. Since we know \(a\) is positive -> \(b\) and \(c\) are positive, too. And that \(0 < c^3 < a < c\) -> \(c <1\).
The first way to think, if you choose a random number \(0<c<1\), (0.3, as it helps you imagine better) you can pick \(b<c\) that the gap is very small (\(10^{-15}\) for example), then \(b^2\approx{c^2}\) (of course \( b^2\) is the smaller one, but the gap is insignificant for our purpose). Compared to this gap, the gap of \(c\) and \(1\) is enormous that \(b^2*1 > c^2*c = c^3\). The same thinking way for \(a\) and \(b\) that you can find \(a\) to satisfy \(a > b^2\). (II) is possible.
Another way for those who still struggle with the aforementioned lines: let pick \(a=0.5\) for example -> \(0.5^3 = 0.125\). We see that \(0.125 < 0.16 = 0.4^2\) (or you can just try the square of a number smaller than \(0.5\)), so we take \(0.4\) as \(b\), then we easily pick \(c\), (\(0.2\) for instance). (II) is possible
III. This case is easier. Take the same \(b,c\) that we chose in (II), together with a very large \(a\) (1,10,100, ...), we get \(a>c>b\)
So (E) is the answer.