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Bunuel
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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.
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We know that

a > \(b ^ 2\) > \(c ^ 3\)

Because \(b ^ 2\) is positive, b can be +ve or -ve

I. a>b>c

a = 0.2
b = 0.35 | \(b ^ 2\) = 0.1425
c = 0.4 | \(c ^ 3\) = 0.096

Valid

II. c>b>a

a = 0.3
b = 0.5 | \(b ^ 2\) = 0.25
c = 0.6 | \(c ^ 3\) = 0.216

Valid

III. a>c>b

a = 0.3
b = -0.5 | \(b ^ 2\) = 0.25
c = -0.3 | \(c ^ 3\) = -0.027

Valid

IMO E
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If a>b^2>c^3, which of the following may be true?

I. a>b>c
II. c>b>a
III. a>c>b

Consider
a = 3, b = 4, c = 5
3<16<25
a<b<c - II. c>b>a

a = 0.3, b = 0.4, c = 0.5
0.3>0.16>0.125
a>b>c - I a>b>c

a = -3, b = -4, c = -5
-125<-3 < 16 - c<a<b

Correct Option C
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[quote="Bunuel"]If \(a > b^2 > c^3\), which of the following may be true?

I. \(a > b > c\)
II. \(c > b > a\)
III. \(a > c > b\)

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III


Let, a =8, b=2, c=1... a>b^2>c^3 satisfies...
Then, a>b>c.. I. correct.
Let, a=1.5, b =-1, c=0.5..a>b^2>c^3 satisfies
The, a>c>b....III Correct

But to satisfy the given condition, a cannot be the lowest.

IMO Answer D


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