Okay, so I need to find the greatest possible value of the expression \(\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)\) where \(a\) is between \(-6\) and \(-2\) and \(b\) is between \(3\) and \(5\). Hmm, let's start by understanding the problem step by step.

First, let me write down the expression again to make sure I have it right: \(\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)\). This looks like a product of two binomials. Maybe I can expand this expression to simplify it. Let me try that.

Multiplying the two terms:

\[
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = a \cdot \frac{1}{b} - a^2 + \frac{1}{b} \cdot \frac{1}{b} - \frac{1}{b} \cdot a
\]

Wait, let me check that again. When you multiply two binomials, it's like (x + y)(z - x) where x = a, y = 1/b, and z = 1/b. Wait, no, actually, the second term is subtracting a, so it's (x + y)(z - x). Let me do this step by step.

First term: \(a \cdot \frac{1}{b}\) which is \(\frac{a}{b}\).

Second term: \(a \cdot (-a)\) which is \(-a^2\).

Third term: \(\frac{1}{b} \cdot \frac{1}{b}\) which is \(\frac{1}{b^2}\).

Fourth term: \(\frac{1}{b} \cdot (-a)\) which is \(-\frac{a}{b}\).

So combining all these terms:

\(\frac{a}{b} - a^2 + \frac{1}{b^2} - \frac{a}{b}\)

Wait, look at that. The \(\frac{a}{b}\) and \(-\frac{a}{b}\) terms cancel each other out. So the entire expression simplifies to:

\(-a^2 + \frac{1}{b^2}\)

Oh, that's much simpler! So instead of dealing with the product of two binomials, it's just \(-a^2 + \frac{1}{b^2}\). That makes things easier. So now the problem reduces to maximizing \(-a^2 + \frac{1}{b^2}\) given the ranges for \(a\) and \(b\).

Alright, so the expression is \(-a^2 + \frac{1}{b^2}\). Let's analyze how this behaves with respect to \(a\) and \(b\).

First, let's consider the term \(-a^2\). Since \(a\) is negative (between \(-6\) and \(-2\)), squaring it will make it positive, and then the negative sign in front will make it negative. So \(-a^2\) is always negative. To maximize the entire expression, we need to minimize \(-a^2\), which is equivalent to making \(a^2\) as small as possible. But since \(a^2\) is squared, the smallest value occurs when \(a\) is closest to zero. However, \(a\) is negative, so the smallest \(a^2\) occurs when \(a\) is closest to zero, i.e., when \(a\) is at its maximum value, which is \(-2\). Wait, but \(-a^2\) would be maximized when \(a^2\) is minimized. Let me confirm that.

Suppose \(a\) is between \(-6\) and \(-2\). The value of \(a^2\) is \(36\) when \(a = -6\), and \(4\) when \(a = -2\). So \(a^2\) is minimized when \(a = -2\), so \(-a^2\) is maximized when \(a = -2\) because \(-4\) is greater than \(-36\). So to maximize the expression \(-a^2 + \frac{1}{b^2}\), we want to choose the largest possible \(-a^2\), which occurs at \(a = -2\). Alternatively, the smaller \(a^2\) is, the larger \(-a^2\) becomes. So yes, \(a = -2\) gives the maximum for the \(-a^2\) term.

Now, looking at the other term, \(\frac{1}{b^2}\). To maximize the entire expression, we need to maximize \(\frac{1}{b^2}\) as well. Since \(b\) is between \(3\) and \(5\), \(b^2\) is between \(9\) and \(25\), so \(\frac{1}{b^2}\) is between \(\frac{1}{25}\) and \(\frac{1}{9}\). To maximize \(\frac{1}{b^2}\), we need the smallest \(b^2\), which occurs when \(b\) is as small as possible, i.e., \(b = 3\). Therefore, \(\frac{1}{b^2}\) is maximized at \(b = 3\), giving \(\frac{1}{9}\).

Therefore, combining both observations, the maximum value of the expression \(-a^2 + \frac{1}{b^2}\) is achieved when \(a = -2\) and \(b = 3\). Plugging these values in:

\(-(-2)^2 + \frac{1}{3^2} = -4 + \frac{1}{9} = -\frac{36}{9} + \frac{1}{9} = -\frac{35}{9}\)

Wait, but that's negative. Is that the maximum? Because if we can get a positive number, that would be better. Wait, maybe I made a mistake here.

Wait, hold on. Let me double-check. The original expression after simplification is \(-a^2 + \frac{1}{b^2}\). So if \(a\) is negative, \(a^2\) is positive, so \(-a^2\) is negative. Then adding \(\frac{1}{b^2}\), which is positive. So depending on the values, the total could be positive or negative. But we need the maximum value. So maybe there's a combination where \(-a^2 + \frac{1}{b^2}\) is positive.

Wait, but according to the earlier analysis, the maximum of \(-a^2\) is when \(a^2\) is minimized, which is \(-4\), and the maximum of \(\frac{1}{b^2}\) is \(\frac{1}{9}\). So adding them together gives \(-4 + \frac{1}{9} = -\frac{35}{9}\), which is approximately \(-3.89\). But maybe there's a combination where \(\frac{1}{b^2}\) is larger than \(a^2\), making the total expression positive? Wait, let's think.

Suppose \(a = -2\), then \(-a^2 = -4\). If \(b\) is smaller, say \(b = 3\), \(\frac{1}{9}\) is about \(0.111\), so total is \(-3.888\). If \(b\) is smaller, say \(b = 2\), which is not allowed here because \(b\) is at least 3. So \(b\) can't be 2. Wait, \(b\) is between 3 and 5. So \(b\) can't be smaller than 3. So \(\frac{1}{b^2}\) is at most \(\frac{1}{9}\). So even if we take \(a\) as large as possible (i.e., \(a = -6\)), then \(-a^2 = -36\), but \(\frac{1}{b^2}\) is at most \(\frac{1}{9}\), so the total is \(-36 + \frac{1}{9} = -\frac{323}{9}\), which is more negative. So actually, the maximum is achieved when \(-a^2\) is as large as possible (i.e., \(a = -2\)) and \(\frac{1}{b^2}\) is as large as possible (i.e., \(b = 3\)), which gives \(-4 + \frac{1}{9} = -\frac{35}{9}\). But the problem asks for the greatest possible value, which is the maximum of this expression. However, since this is negative, but maybe there's a way to get a positive value?

Wait, perhaps I made a mistake in expanding the original expression. Let me check that again.

Original expression: \(\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)\)

Multiply these two terms:

First term: \(a \cdot \frac{1}{b} = \frac{a}{b}\)

Second term: \(a \cdot (-a) = -a^2\)

Third term: \(\frac{1}{b} \cdot \frac{1}{b} = \frac{1}{b^2}\)

Fourth term: \(\frac{1}{b} \cdot (-a) = -\frac{a}{b}\)

So combining:

\(\frac{a}{b} - a^2 + \frac{1}{b^2} - \frac{a}{b}\)

The \(\frac{a}{b}\) and \(-\frac{a}{b}\) terms cancel, leaving \(-a^2 + \frac{1}{b^2}\). So that's correct. Therefore, the expression simplifies correctly.

So the expression is \(-a^2 + \frac{1}{b^2}\). Therefore, the maximum value of this expression is when \(-a^2\) is as large as possible and \(\frac{1}{b^2}\) is as large as possible. But since \(-a^2\) is negative, maximizing it (i.e., making it less negative) would require \(a^2\) to be as small as possible. Therefore, \(a = -2\) gives the smallest \(a^2\) (since \(a\) is negative, the square is positive, and the smallest when \(a\) is closest to zero). Then, \(\frac{1}{b^2}\) is maximized when \(b\) is as small as possible, which is \(3\). So plugging in, we get:

\(-(-2)^2 + \frac{1}{3^2} = -4 + \frac{1}{9} = -\frac{36}{9} + \frac{1}{9} = -\frac{35}{9}\)

But wait, this is negative. But the problem is asking for the greatest possible value. If the expression can only be negative, then the maximum is \(-\frac{35}{9}\). However, maybe there's a case where the expression is positive? Let me check.

Suppose \(a\) is negative and \(b\) is in between. Let's pick \(a = -2\), \(b = 3\). Then expression is \(-4 + 1/9 = -35/9 ≈ -3.89\). If I take \(a = -2\), \(b = 3.5\), then \(\frac{1}{b^2} = 1/(12.25) ≈ 0.0816\), so expression is \(-4 + 0.0816 ≈ -3.9184\), which is worse. If I take \(a = -2\), \(b = 3\), then it's the same as before. If I take \(a = -2\), \(b = 3\), that's the maximum.

Alternatively, if I take \(a\) more negative, say \(a = -6\), then \(-a^2 = -36\), and even if \(b = 3\), \(\frac{1}{9}\), so total is \(-36 + 1/9 ≈ -35.888\), which is worse. So indeed, the maximum is at \(a = -2\), \(b = 3\), giving \(-35/9\).

Wait, but the question says "greatest possible value", which could be negative. But maybe I need to check if there's a different combination of \(a\) and \(b\) where the original expression is positive. Let me check with different values.

Suppose \(a = -3\), \(b = 3\). Then original expression:

\(\left(-3 + 1/3\right)\left(1/3 - (-3)\right) = (-8/3)(10/3) = -80/9 ≈ -8.89\)

Negative again. If I take \(a = -2\), \(b = 5\):

\(\left(-2 + 1/5\right)\left(1/5 - (-2)\right) = (-9/5)(11/5) = -99/25 ≈ -3.96\)

Which is slightly worse than -35/9 ≈ -3.89. If I take \(a = -2\), \(b = 3\), which gives -35/9. If I take \(a = -2\), \(b = 4\):

\(\left(-2 + 1/4\right)\left(1/4 - (-2)\right) = (-7/4)(9/4) = -63/16 ≈ -3.9375\)

Which is worse than -35/9. So indeed, the maximum seems to be at \(a = -2\), \(b = 3\), giving -35/9. But since the problem asks for the greatest possible value, which is the maximum of this expression, and since this is the highest value we can get, even though it's negative, that's the answer.

Wait, but let me check another approach. Let me consider the original expression again:

\(\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)\)

Let me denote \(x = a\) and \(y = \frac{1}{b}\). Then the expression becomes:

\((x + y)(y - x) = y^2 - x^2\)

Which is the same as \(-x^2 + y^2\), which matches our previous simplification. So the expression is \(y^2 - x^2\). Since \(a\) is negative, \(x = a\) is negative, \(y = 1/b\) is positive. So \(y^2 - x^2\) is the difference of squares. To maximize this, since \(y^2\) is positive and \(x^2\) is positive, we need to maximize \(y^2\) and minimize \(x^2\). So same conclusion as before: \(x^2\) is minimized when \(x\) is closest to zero (i.e., \(a = -2\)), and \(y^2\) is maximized when \(y\) is as large as possible (i.e., \(b\) as small as possible, \(b = 3\)), giving \(y = 1/3\). Therefore, the maximum is \((1/3)^2 - (-2)^2 = 1/9 - 4 = -35/9\). So that's the same result. Therefore, the answer is \(-35/9\), which as a common fraction is \(-\frac{35}{9}\). But the problem says "greatest possible value", so even though it's negative, that's the maximum.

Wait, but the original problem didn't specify that the answer has to be positive. So if the maximum value is negative, then that's the answer. But let me check once more. Suppose I take \(a = -2\), \(b = 3\), the expression is \(-35/9\). If I take \(a\) slightly more negative, say \(a = -1.9\), which is still between -6 and -2? Wait, no, -1.9 is greater than -2, so it's not in the allowed range. So \(a\) has to be between -6 and -2, so the smallest possible \(a\) is -6, the largest is -2. So \(a = -2\) is the maximum \(a\). So that's correct. Then \(b = 3\) is the minimum \(b\), giving the maximum \(\frac{1}{b^2}\). Therefore, the conclusion is correct.

Therefore, the greatest possible value is \(-35/9\), which is approximately -3.89. But since the problem asks for a common fraction, the answer is \(-\frac{35}{9}\). However, the problem says "greatest possible value", which could imply the largest value, regardless of sign. So even though it's negative, it is the maximum possible. Therefore, the answer is \(-\frac{35}{9}\).

Wait, but let me check once more. Suppose someone argues that maybe there's a combination where the original expression is positive. Let's see. If \(\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)\) is positive, then the product of the two terms must be positive. That happens when both terms are positive or both are negative.

First case: both terms positive. Then \(a + \frac{1}{b} > 0\) and \(\frac{1}{b} - a > 0\). Let's see:

\(a + \frac{1}{b} > 0\): Since \(a\) is negative and \(\frac{1}{b}\) is positive, this would require that \(\frac{1}{b} > |a|\). Similarly, \(\frac{1}{b} - a > 0\): Since \(a\) is negative, subtracting a negative is adding, so this is always true because \(\frac{1}{b}\) is positive and \( -a \) is positive. So the second term is always positive. Therefore, for both terms to be positive, we need \(a + \frac{1}{b} > 0\). So \(\frac{1}{b} > |a|\). But \(a\) is at least -6, so \(|a|\) is at most 6. But \(\frac{1}{b}\) is at most 1/3 ≈ 0.333. Since 0.333 < 6, this inequality can never be satisfied. Therefore, both terms cannot be positive. Therefore, the original expression can't be positive. Therefore, the expression is always negative or zero. Wait, when can it be zero?

When either \(a + 1/b = 0\) or \(1/b - a = 0\).

Case 1: \(a + 1/b = 0\). Then \(a = -1/b\). Given that \(a\) is between -6 and -2, and \(b\) is between 3 and 5, \(1/b\) is between 1/5 and 1/3. So \(a = -1/b\) would require \(a\) between -1/3 and -1/5. But our \(a\) is between -6 and -2, so this is impossible. Therefore, no solution here.

Case 2: \(1/b - a = 0\). Then \(a = 1/b\). But \(a\) is negative, and \(1/b\) is positive. Therefore, \(a = 1/b\) is impossible. Therefore, the expression can never be zero. Therefore, the expression is always negative. Therefore, the maximum value is indeed negative, and the maximum is \(-35/9\). Therefore, the answer is \(-\frac{35}{9}\).

But since the problem asks for the greatest possible value, which is the maximum value, even if it's negative, the answer is \(-\frac{35}{9}\). Therefore, I think that's the final answer.
Therefore, the greatest possible value of the given expression is $\boxed{-\frac{35}{9}}$.
</think>