Explanation:
Okay, so I need to find the greatest possible value of the expression $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$ given the constraints $-6 \leq a \leq -2$ and $3 \leq b \leq 5$. Hmm, let me start by understanding what this expression is and how I can simplify or analyze it.

First, maybe I should try expanding the expression to see if it can be simplified. Let me do that:

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

This looks like a product of two terms. Let me multiply them out. Remembering the distributive property, which is like FOIL for binomials:

First terms: $a \times \frac{1}{b} = \frac{a}{b}$

Outer terms: $a \times (-a) = -a^2$

Inner terms: $\frac{1}{b} \times \frac{1}{b} = \frac{1}{b^2}$

Last terms: $\frac{1}{b} \times (-a) = -\frac{a}{b}$

Now combine all these terms:

$$
\frac{a}{b} - a^2 + \frac{1}{b^2} - \frac{a}{b}
$$

Wait, the first and the last terms are $\frac{a}{b}$ and $-\frac{a}{b}$, which cancel each other out! So that simplifies the expression to:

$$
- a^2 + \frac{1}{b^2}
$$

Oh, that's much simpler! So the original expression simplifies to $-a^2 + \frac{1}{b^2}$. Therefore, the problem reduces to maximizing $-a^2 + \frac{1}{b^2}$ given the constraints on $a$ and $b$.

Alright, so now I need to maximize $-a^2 + \frac{1}{b^2}$. Let me think about how to approach this. Since $a$ and $b$ are independent variables (their ranges don't overlap or interact except through the expression), maybe I can analyze each term separately?

But wait, actually, the expression is a combination of a function of $a$ and a function of $b$. So, since they are separate variables, maybe the maximum occurs when each term is maximized? However, I need to check if increasing one term would affect the other. Let me think.

But actually, since $-a^2$ is a function of $a$ only, and $\frac{1}{b^2}$ is a function of $b$ only, their combination is the sum of two separate functions. So, if I want to maximize the entire expression, I can maximize each term individually and then add them up. But I need to verify if that's valid here.

Wait, but suppose that maximizing $-a^2$ and maximizing $\frac{1}{b^2}$ are independent. Since $a$ and $b$ are independent variables, then yes, the maximum of the sum is the sum of the maxima. However, let me confirm this logic.

Suppose that for $-a^2$, since $a$ is between -6 and -2, then $a^2$ is between 4 and 36. Therefore, $-a^2$ is between -36 and -4. So, the maximum of $-a^2$ is -4, which occurs when $a^2$ is minimized, i.e., when $a$ is as close to zero as possible. Since $a$ is between -6 and -2, the smallest $a^2$ is when $a = -2$, because (-2)^2 = 4, which is less than (-6)^2 = 36. So maximum of $-a^2$ is -4 when a = -2.

Similarly, for $\frac{1}{b^2}$, since $b$ is between 3 and 5, $b^2$ is between 9 and 25, so $\frac{1}{b^2}$ is between 1/25 and 1/9. Therefore, maximum of $\frac{1}{b^2}$ is 1/9 when $b$ is minimized, i.e., when $b = 3$.

Therefore, if I can take $a = -2$ and $b = 3$, then the expression would be $-4 + 1/9 = -4 + 1/9 = -35/9$. But wait, is this the maximum?

But hold on, maybe there's a combination where even though $-a^2$ is not maximized, but $\frac{1}{b^2}$ is larger? Or vice versa? Wait, but since $-a^2$ is always negative, and $\frac{1}{b^2}$ is always positive. So to maximize the total expression, we need to maximize the positive term and minimize the negative term. Because $-a^2$ is negative, so to make the total expression as large as possible, we need to make it as close to zero as possible (i.e., minimize the negative part) and maximize the positive part.

So, that would mean:

Maximize $\frac{1}{b^2}$ (which is achieved when b is as small as possible, i.e., b = 3) and minimize $-a^2$ (which is equivalent to maximizing $a^2$, because $-a^2$ is negative, so minimizing it would mean making it more negative, but wait that contradicts.

Wait, hold on, maybe my earlier conclusion was wrong.

Wait, the expression is $-a^2 + \frac{1}{b^2}$. So, if I think of this as ($-a^2$) + ($\frac{1}{b^2}$). Therefore, to maximize this, I need to maximize each term. However, $-a^2$ is a negative term, so maximizing it would mean making it less negative, i.e., closer to zero. Whereas $\frac{1}{b^2}$ is positive, so maximizing it would make that term larger. Therefore, the maximum occurs when both terms are maximized. But how?

Wait, so maximizing $-a^2$ would require minimizing $a^2$, which occurs at $a = -2$ (since $a$ is negative). Then, $\frac{1}{b^2}$ is maximized when $b$ is at its minimum, which is 3. So combining these, with $a = -2$ and $b = 3$, gives us the maximum value of the expression. Which is $-(-2)^2 + (1/3)^2 = -4 + 1/9 = -35/9$. But this is negative. Wait, is this the maximum?

Wait, maybe I need to check if there are other possibilities where even though one term is smaller, the combination is larger? For example, suppose that increasing $a^2$ (so making $-a^2$ more negative) but increasing $\frac{1}{b^2}$ enough to compensate? But since $\frac{1}{b^2}$ is positive and $-a^2$ is negative, how can increasing $\frac{1}{b^2}$ compensate for a more negative term?

Wait, for example, suppose that I take $a = -6$ and $b = 5$. Then the expression would be $-(-6)^2 + (1/5)^2 = -36 + 1/25 = -35.96$, which is way smaller than -35/9 ≈ -3.888... So that's worse. If I take $a = -2$ and $b = 5$, then it's $-4 + 1/25 = -3.96$, which is still worse than -35/9 ≈ -3.888...

Wait, but if I take $a = -2$ and $b = 3$, that gives me -4 + 1/9 ≈ -3.888..., which is better. Wait, let me check if there's a way to get a higher value (i.e., less negative) by combining different a and b values?

Alternatively, maybe my initial approach is wrong? Let me check again.

Wait, the original expression, after simplifying, is $-a^2 + \frac{1}{b^2}$. So, in order to maximize this expression, since $-a^2$ is negative and $\frac{1}{b^2}$ is positive, we need to make $-a^2$ as small as possible (i.e., as close to zero as possible) and $\frac{1}{b^2}$ as large as possible. Therefore, to minimize $-a^2$, we need to maximize $a^2$. Wait, hold on, because if you have $-a^2$, then minimizing $-a^2$ would mean making $-a^2$ as small as possible, but since $a^2$ is positive, $-a^2$ is negative. Therefore, the smallest value of $-a^2$ is when $a^2$ is largest. Wait, but that's contradictory with the earlier thought. Wait, no:

Wait, suppose I have two numbers, say, -5 and -3. Which one is smaller? -5 is smaller than -3. So, if I want to minimize $-a^2$, I need to make $-a^2$ as small as possible, which would be when $a^2$ is as large as possible. Therefore, the minimum of $-a^2$ is when $a^2$ is maximum. Therefore, if I want to minimize $-a^2$, which is equivalent to maximizing $a^2$. Therefore, in order to have the expression $-a^2 + \frac{1}{b^2}$ as large as possible, since $-a^2$ is being minimized (i.e., going more negative) and $\frac{1}{b^2}$ is being maximized, but does that help?

Wait, hold on. Let me think again. If you have an expression like $X + Y$, and you want to maximize it. If X is negative and Y is positive, you need to make X as large as possible (i.e., closer to zero) and Y as large as possible. Because making X larger (less negative) and Y larger would add up to a larger total. Therefore, to maximize $X + Y$, with $X = -a^2$ and $Y = \frac{1}{b^2}$, you need to maximize $X$ (i.e., make $-a^2$ as large as possible, which is equivalent to making $a^2$ as small as possible) and maximize $Y$ (i.e., make $b$ as small as possible). Therefore, that would be the case when $a$ is as close to zero as possible (i.e., $a = -2$) and $b$ as small as possible (i.e., $b = 3$). Therefore, the maximum value is $-(-2)^2 + (1/3)^2 = -4 + 1/9 = -35/9$, which is approximately -3.888...

But wait, is there a way to get a higher value (i.e., a less negative number) by choosing different a and b?

Wait, let me check with other values. For example, suppose I choose $a = -2$ and $b = 3$, as before, giving $-4 + 1/9 = -35/9 ≈ -3.888$. What if I choose $a = -3$ and $b = 3$? Then the expression would be $-(-3)^2 + 1/9 = -9 + 1/9 = -80/9 ≈ -8.888$, which is worse. If I take $a = -2$ and $b = 5$, then it's $-4 + 1/25 = -4 + 0.04 = -3.96$, which is worse than -35/9. If I take $a = -2$ and $b = 3$, that's -35/9 ≈ -3.888...

Wait, what if I take $a = -2$ and $b = something else$? Let me think. Since $\frac{1}{b^2}$ is maximized at $b = 3$, so that's the best possible for that term. So, perhaps that is the maximum? But maybe if I take some other combination where the negative term is less negative and the positive term is not as large? Wait, but since we can't increase the positive term beyond 1/9, and we can't make the negative term less negative than -4 (since $-a^2$ is minimized at -4 when $a = -2$), then maybe that is indeed the maximum. Wait, but hold on, let me check if there's some combination where $-a^2 + \frac{1}{b^2}$ is larger than -35/9. Let's try.

Suppose we take $a = -2$ and $b = 3$, then value is -35/9 ≈ -3.888. Let me try $a = -2.5$ and $b = 3$. Then, $-a^2 = -(-2.5)^2 = -6.25$, and $\frac{1}{b^2} = 1/9$. So total is -6.25 + 1/9 ≈ -6.25 + 0.111 ≈ -6.139, which is worse. So worse.

If I take $a = -1.5$, but wait a can't be -1.5 because a is between -6 and -2. So the minimum value of a is -6 and maximum is -2. So $a$ can't be between -2 and 0. Therefore, $a$ is always negative, between -6 and -2. So the closest it can get to zero is -2. Therefore, if I take $a = -2$, that's the maximum value (closest to zero) of a. Therefore, $-a^2$ is maximized at $a = -2$ (since $-a^2$ is maximum when $a^2$ is minimum). Therefore, yes, so with that, if I also take $b = 3$, then the expression is maximized. Therefore, the maximum value is -35/9? Wait, but that seems counterintuitive because the original expression was $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$, which we simplified to $-a^2 + \frac{1}{b^2}$. But if that's the case, then maybe there's a mistake in simplification?

Wait, let me double-check the expansion. Let me redo the multiplication step by step.

Original expression:

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

Let me consider this as $(X + Y)(Y - X)$ where $X = a$ and $Y = \frac{1}{b}$. Wait, that's similar to $(Y + X)(Y - X) = Y^2 - X^2$. Wait, that's a difference of squares! So:

$$
(Y + X)(Y - X) = Y^2 - X^2
$$

Therefore, substituting back:

$$
\left(\frac{1}{b}\right)^2 - a^2 = \frac{1}{b^2} - a^2
$$

Which is the same as $-a^2 + \frac{1}{b^2}$. So that's correct. So the simplification is correct.

So then, the expression is indeed equal to $\frac{1}{b^2} - a^2$. Therefore, to maximize this, we need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Which means, maximize $\frac{1}{b^2}$ by taking the smallest possible b (which is 3) and minimize $a^2$ by taking the value of $a$ closest to 0, which is $a = -2$.

Therefore, the maximum value is $\frac{1}{3^2} - (-2)^2 = \frac{1}{9} - 4 = -\frac{35}{9}$. But this is negative. Wait, but can we get a positive value?

Wait, if the original expression was $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$, is there any chance that it can be positive? Let me check with specific values.

Take $a = -2$ and $b = 3$:

First term: $-2 + 1/3 ≈ -2 + 0.333 ≈ -1.666$

Second term: $1/3 - (-2) = 1/3 + 2 ≈ 2.333$

Multiplying them: (-1.666)(2.333) ≈ -3.888, which is the same as -35/9 ≈ -3.888...

What if I take a different combination? Let's say $a = -2$ and $b = 5$:

First term: -2 + 1/5 = -1.8

Second term: 1/5 - (-2) = 1/5 + 2 = 2.2

Product: (-1.8)(2.2) = -3.96, which is worse. So still negative.

What if I take $a = -3$ and $b = 3$:

First term: -3 + 1/3 ≈ -2.666

Second term: 1/3 - (-3) ≈ 1/3 + 3 ≈ 3.333

Product ≈ (-2.666)(3.333) ≈ -8.888...

Still negative.

Wait, so is there any possible combination where the product is positive?

Let me think. For the product $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$ to be positive, either both factors are positive or both are negative.

Let me analyze the signs of each term.

First, since $a$ is between -6 and -2, so $a$ is negative. $b$ is between 3 and 5, so $1/b$ is between 1/5 and 1/3, so positive. Therefore, $a + 1/b$ is negative plus positive. Depending on the values, this could be negative or positive?

Similarly, $\frac{1}{b} - a$ is positive minus negative, which is positive + positive, so positive. Therefore, the second term is always positive. Therefore, the product is positive times (either positive or negative). So for the entire product to be positive, the first term must also be positive. Therefore, $\left(a + \frac{1}{b}\right)$ must be positive and $\left(\frac{1}{b} - a\right)$ is always positive. Therefore, the entire expression is positive if and only if $a + \frac{1}{b} > 0$, which is equivalent to $a > -\frac{1}{b}$. Since $a$ is negative and $b$ is positive, this inequality may or may not hold.

So, if there exists values of $a$ and $b$ such that $a > -1/b$, then the product is positive. Therefore, maybe there is a maximum positive value?

Wait, this is an important point. Earlier, I thought the expression was always negative, but actually, it can be positive if $a + 1/b$ is positive. Therefore, my previous conclusion that the maximum value is -35/9 is wrong because it was assuming that the maximum occurs when the expression is negative. However, there might be positive values, which could be larger.

Therefore, I need to re-examine the problem. So the expression can be positive or negative. Therefore, the maximum value could be positive. Therefore, I need to consider both possibilities.

Let me first check whether $a + 1/b$ can be positive. That would mean $a > -1/b$. Since $a$ is between -6 and -2, and $b$ is between 3 and 5. Therefore, $-1/b$ is between -1/5 and -1/3. So, for $a > -1/b$, given that $a$ is between -6 and -2, we need to find if there exists $a$ and $b$ such that $a > -1/b$.

Let me think. Since $a$ is at least -6 and at most -2, and $-1/b$ is between -1/5 and -1/3 ≈ -0.333. Since $a$ is at least -6, but to have $a > -1/b$, which is greater than -1/3. So, if $a$ is between -2 and -1/3, then $a$ is greater than -1/b (since $b$ is at least 3, so $-1/b$ is at most -1/3). However, our constraint is $a$ between -6 and -2, so $a$ is between -6 and -2. Therefore, $a$ is always less than or equal to -2, which is less than -1/3. Wait, because -2 is approximately -2.0, which is less than -0.333. Therefore, $a$ is always less than -1/b? Wait, no. Wait, if $a$ is between -6 and -2, and $-1/b$ is between -1/5 and -1/3. For example, take $b = 5$, so $-1/b = -1/5 = -0.2$. So, if $a$ is between -6 and -2, then $a$ is less than -0.2. Therefore, $a + 1/b$ is $a + 1/b$, where $a < -0.2$ and $1/b < 0.2$. Therefore, adding them, $a + 1/b < -0.2 + 0.2 = 0$, but how much less?

Wait, let's suppose $a = -2$, $b = 5$. Then $a + 1/b = -2 + 0.2 = -1.8 < 0$. If $a = -2$, $b = 3$, $a + 1/b = -2 + 1/3 ≈ -1.666 < 0$. If $a = -2.5$, $b = 5$: $-2.5 + 0.2 = -2.3 < 0$. If $a = -1.5$, but $a$ can't be -1.5. Wait, so even if $b$ is as small as 3, then $-1/b = -1/3 ≈ -0.333$. So, if $a$ is between -6 and -2, then $a$ is always less than -0.333? Wait, no. Wait, if $a$ is between -6 and -2, then the maximum value of $a$ is -2, which is less than -1/3 ≈ -0.333? Wait no! Wait, -2 is less than -1/3? Yes, because -2 is further to the left on the number line. So $a$ is always less than -1/3, which is the upper bound of $-1/b$. Therefore, $a + 1/b < -1/b + 1/b = 0$. Therefore, $a + 1/b$ is negative for all allowed $a$ and $b$. Therefore, the first term is negative, and the second term is positive, so the entire product is negative. Therefore, the expression is always negative. Therefore, my initial conclusion that the maximum value is -35/9 is correct? But then why did I get confused earlier about positive values? Because I thought maybe there could be positive values, but actually, given the constraints, it's always negative.

Therefore, the expression is always negative, so the maximum possible value is the one closest to zero, which is achieved when the expression is least negative. Which would be when $-a^2 + \frac{1}{b^2}$ is closest to zero, i.e., when $-a^2 + \frac{1}{b^2}$ is as large as possible (since it's negative).

So, as before, to maximize $-a^2 + \frac{1}{b^2}$, which is equivalent to minimizing $a^2$ and maximizing $\frac{1}{b^2}$. Therefore, as before, take $a = -2$ (minimizing $a^2$) and $b = 3$ (maximizing $\frac{1}{b^2}$), which gives the value of $-4 + 1/9 = -35/9$. But I should check if there's another combination where the value is higher (i.e., less negative) than -35/9.

Wait, for instance, suppose I take $a = -2$ and $b = 3$: value is -35/9 ≈ -3.888...

What if I take $a = -2$ and $b$ slightly larger? Let's take $b = 4$. Then the value is $-(-2)^2 + 1/16 = -4 + 1/16 = -63/16 ≈ -3.9375$, which is worse (more negative). Similarly, if I take $b = 3.5$, then $\frac{1}{b^2} ≈ 1/(12.25) ≈ 0.0816$, so total value ≈ -4 + 0.0816 ≈ -3.9184, which is worse than -35/9 ≈ -3.888... So worse.

Alternatively, if I take $a$ slightly greater than -2 (i.e., closer to zero), but $a$ can't be greater than -2 because the upper bound is -2. So $a = -2$ is the closest to zero. What if I take $a = -2$ and $b = 3$, which gives -35/9. What if I take $a$ slightly less than -2, like $a = -2.1$ (but since $a$ can be as low as -6, but let's take a value between -6 and -2). Wait, if I take $a$ further from zero, then $-a^2$ becomes more negative. So that would make the entire expression more negative. Therefore, worse. Therefore, the maximum value is indeed achieved at $a = -2$ and $b = 3$.

Wait, but wait, there's another thought. Maybe if I take a different value of $a$ and $b$ such that $-a^2 + \frac{1}{b^2}$ is larger? For example, maybe there's a balance between decreasing $a^2$ and increasing $\frac{1}{b^2}$?

Wait, for example, suppose $a = -2$, which gives $-a^2 = -4$, and $b = 3$ gives $\frac{1}{b^2} = 1/9$. Total is -35/9 ≈ -3.888...

Alternatively, suppose I take $a = -\sqrt{\frac{1}{b^2}}$... Wait, but that's not helpful. Maybe think of the expression as a function of two variables and use calculus to find maxima?

But since this is a problem likely intended for algebra, maybe calculus is not expected. However, let's try. Maybe taking partial derivatives?

But let me think again. Since the expression is $-a^2 + \frac{1}{b^2}$, with constraints on $a$ and $b$. So to maximize this expression, we can consider that $-a^2$ is maximized when $a^2$ is minimized, which is when $a$ is closest to zero, i.e., $a = -2$. Similarly, $\frac{1}{b^2}$ is maximized when $b$ is minimized, i.e., $b = 3$. Therefore, the combination of these two gives the maximum value of the expression. Therefore, the answer is -35/9. But I need to check again if there's any other combination where the value is higher.

Wait, suppose I take $a = -2$ and $b = 3$, giving -35/9. Suppose I take $a = -2$ and $b = something else$. For example, take $b = 4$. Then the expression is -4 + 1/16 = -63/16 = -3.9375. Which is less than -35/9 (-35/9 ≈ -3.888). Similarly, if I take $a = -2$ and $b = 5$, it's -4 + 1/25 = -4 + 0.04 = -3.96, which is worse.

Alternatively, take $a$ slightly different. For example, take $a = -2$ and $b = 3$, and compare to $a = -2.5$ and $b = something$. Let me try $a = -2.5$, and find a $b$ such that maybe the expression is better. Let's compute $-a^2 + 1/b^2 = -6.25 + 1/b^2$. To make this as large as possible, we need to maximize $1/b^2$, which is when $b$ is minimized (3). Therefore, the value would be -6.25 + 1/9 ≈ -6.25 + 0.111 ≈ -6.139, which is worse than -3.888. Therefore, worse.

Alternatively, suppose I take $a = -2$ and $b = 3$, which is the best. So maybe that is indeed the maximum. Therefore, the answer is -35/9. But wait, the problem says "the greatest possible value". Since the expression is always negative, the greatest possible value is the one closest to zero, which is -35/9. But let me check one more time with different a and b values.

Suppose $a = -2$ and $b = 3$, expression is -35/9 ≈ -3.888...

What if $a = -2$ and $b = \sqrt{1/(something)}$... Wait, but $b$ is between 3 and 5, so no. Alternatively, if there is some $a$ and $b$ where the two terms balance out to give a higher value. For example, maybe $-a^2 + 1/b^2 = 0$? That would require $1/b^2 = a^2$, so $a = \pm 1/b$. However, since $a$ is negative and $1/b$ is positive, $a = -1/b$. So $a = -1/b$. Is there a $a$ and $b$ in the given ranges where this holds?

Let me solve for $a = -1/b$. Given $a$ is between -6 and -2, $b$ is between 3 and 5. Then $a = -1/b$ implies that $ -6 \leq -1/b \leq -2 $. Multiply all parts by -1 (and reverse inequalities):

$6 \geq 1/b \geq 2$, which is equivalent to $2 \leq 1/b \leq 6$. Taking reciprocals (and reversing inequalities again because all are positive):

$1/6 \leq b \leq 1/2$. But $b$ is between 3 and 5, so this is impossible. Therefore, there is no solution where $a = -1/b$ within the given ranges. Therefore, the expression cannot be zero. Therefore, the expression is always negative, so the maximum value is indeed -35/9.

But wait, hold on. Wait, maybe I can check with another approach. Let me think of the original expression before simplifying, $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me consider substituting variables. Let me denote $x = a$ and $y = \frac{1}{b}$. Then, since $a$ is between -6 and -2, $x$ is between -6 and -2. Since $b$ is between 3 and 5, $y$ is between 1/5 and 1/3.

Then the expression becomes $(x + y)(y - x) = y^2 - x^2$. So again, same result as before. So we need to maximize $y^2 - x^2$ given that $x \in [-6, -2]$ and $y \in [1/5, 1/3]$.

Therefore, since $y^2$ is maximized when $y$ is as large as possible (i.e., $y = 1/3$) and $x^2$ is minimized when $x$ is as close to zero as possible (i.e., $x = -2$), then the maximum of $y^2 - x^2$ is indeed $(1/3)^2 - (-2)^2 = 1/9 - 4 = -35/9$. So same result.

Therefore, this seems to confirm that -35/9 is the maximum value.

Wait, but the problem says "the greatest possible value". Since all values are negative, the greatest possible value is the one closest to zero, which is indeed -35/9. But maybe I need to check if there's a case where $y^2 - x^2$ is greater than -35/9? For example, take $x = -2$ and $y = 1/3$: -35/9 ≈ -3.888...

If I take $x = -2$ and $y = 1/3$: same.

If I take $x = -2$ and $y = 1/3$, then it's maximum. Any other combination?

Take $x = -2$ and $y = something less than 1/3$, then $y^2$ decreases, so the value decreases. If I take $x$ more negative, then $x^2$ increases, so $y^2 - x^2$ decreases. Therefore, the maximum occurs at $x = -2$, $y = 1/3$. Therefore, the answer is -35/9.

But wait, the problem says "the greatest possible value", and asks to express it as a common fraction. -35/9 is a common fraction. However, is there a possibility that I made a mistake in the sign?

Wait, let me check once again. Original expression:

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

Expanding gives $\frac{1}{b^2} - a^2$, which is the same as $-a^2 + \frac{1}{b^2}$. So yes.

But let me compute with actual numbers. Take $a = -2$, $b = 3$:

First factor: $-2 + 1/3 = -5/3$

Second factor: $1/3 - (-2) = 1/3 + 2 = 7/3$

Product: (-5/3)(7/3) = -35/9. Correct. Therefore, the value is indeed -35/9.

But is there a possibility that the problem expects a positive answer? Maybe I misread the problem? Let me check the original problem again:

"If $-6 \leq a \leq -2$ and $3 \leq b \leq 5$, what is the greatest possible value of $\left(a+\frac{1}{b}\right)\left(\frac{1}{b}-a\right)$? Express your answer as a common fraction."

No, it says "greatest possible value", which is the maximum value. Since all values are negative, the maximum is the least negative one, which is -35/9. Therefore, unless there's a miscalculation, that should be the answer.

Wait, but maybe I need to check if there's another way to interpret the original expression. Maybe there was a typo in my expansion?

Wait, original expression: $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me multiply it again:

First, multiply $a$ by $\frac{1}{b}$: $a/b$

Then, $a$ times $-a$: $-a^2$

Then, $\frac{1}{b}$ times $\frac{1}{b}$: $1/b^2$

Then, $\frac{1}{b}$ times $-a$: $-a/b$

So altogether:

$a/b - a^2 + 1/b^2 - a/b$

Yes, the $a/b$ and $-a/b$ cancel, leaving $-a^2 + 1/b^2$, which is correct. So that's right.

Another way: note that $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2$, which is difference of squares. So that's correct.

Therefore, I think the answer is indeed -35/9.

But just to be thorough, let me check if there's any other combination where $-a^2 + \frac{1}{b^2}$ is larger. For example, take $a = -2$ and $b = 3$, as before: -4 + 1/9 = -35/9 ≈ -3.888...

What if $a = -2$ and $b = 3. Let me take $a$ slightly different. Suppose $a = -2$ and $b$ is variable. Since $-a^2$ is fixed at -4, and $\frac{1}{b^2}$ is maximum at 1/9, so that's the best possible for that combination. If I take $a$ varying, but keeping $a$ at -2 is optimal. So no.

Alternatively, maybe take $a$ slightly different and $b$ slightly different such that the decrease in $-a^2$ is offset by an increase in $\frac{1}{b^2}$. For example, suppose I take $a = -2. Let me pick a $a$ slightly larger than -2 (i.e., closer to zero) but since the maximum $a$ is -2, I can't go any closer. So no. If I take $a$ slightly less than -2 (i.e., further from zero), which would decrease $-a^2$, making it more negative. Therefore, worse.

Alternatively, if I take $b$ slightly smaller than 3, but since $b$ can't be smaller than 3, as given $3 \leq b \leq 5$, so no.

Therefore, there's no way to get a higher value than -35/9.

Wait, but just to check, maybe if $a$ and $b$ are chosen such that $-a^2 + \frac{1}{b^2}$ is larger? Let me think of extremes. For example, when $a = -6$, $b = 5$: $-36 + 1/25 = -35.96$, which is worse. When $a = -2$, $b = 5$: -4 + 1/25 ≈ -3.96. When $a = -2$, $b = 3$: -35/9 ≈ -3.888... So indeed, the latter is higher.

Alternatively, take $a = -2$, $b = 3$: this is the best. Therefore, I think I can be confident that -35/9 is the correct answer.

But just to check with another method. Let me consider that the expression is $-a^2 + \frac{1}{b^2}$. Since both $a$ and $b$ are independent variables, the maximum occurs at the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$.

Since $a$ is between -6 and -2, $a^2$ is between 4 and 36. Therefore, the minimum $a^2$ is 4, achieved at $a = -2$. The maximum $\frac{1}{b^2}$ is 1/9, achieved at $b = 3$. Therefore, maximum value is 1/9 - 4 = -35/9. So same answer.

Therefore, I think this confirms that the answer is indeed -35/9.

But the problem says "the greatest possible value", and the answer is negative. However, maybe I made a mistake in simplifying the expression? Wait, let me check once again with specific numbers. Let me take $a = -2$, $b = 3$. Then the original expression:

First factor: $-2 + 1/3 = -5/3$

Second factor: $1/3 - (-2) = 1/3 + 2 = 7/3$

Product: (-5/3)(7/3) = -35/9. Correct.

If I take $a = -2$, $b = 3$, the value is -35/9. If I take $a = -2$, $b = 3$, is that within the given ranges? Yes, since $-6 \leq -2 \leq -2$ and $3 \leq 3 \leq 5$. So yes.

Therefore, the answer is -35/9. But since the problem says "greatest possible value", and -35/9 is indeed the maximum (since all other combinations give smaller values), then I think that is the answer.

However, just to be 100% sure, let me check another combination where maybe the product is higher. Let me take $a = -2$, $b = 3$: -35/9. What if I take $a = -2$, $b$ approaching 3 from above? Then $1/b^2$ approaches 1/9 from below, so the expression approaches -35/9 from below. So worse. If I take $a$ approaching -2 from below (i.e., more negative), then $-a^2$ becomes more negative, so the expression becomes worse. Therefore, no.

Alternatively, take $a = -2$ and $b = 3$: maximum.

Therefore, conclusion: the greatest possible value is -35/9, which is a common fraction.

But wait, the problem might expect a positive answer? Let me check once again for any possible mistake. Wait, could there be a miscalculation in the expansion?

Original expression:

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

Let me compute this expression with $a = -2$, $b = 3$:

First factor: $-2 + 1/3 = -5/3$

Second factor: $1/3 - (-2) = 1/3 + 2 = 7/3$

Product: $(-5/3)(7/3) = -35/9$. Correct.

Alternatively, if I consider $\frac{1}{b^2} - a^2$ with the same values: $1/9 - 4 = -35/9$. Correct.

Therefore, I think the answer is correct. Therefore, the greatest possible value is $\boxed{-\dfrac{35}{9}}$.

Wait, but the problem says "greatest possible value", and if all values are negative, then the greatest is the one closest to zero. However, sometimes problems might have a trick where the expression can be positive. But according to all the above, it can't. Therefore, unless there's a mistake in my analysis, I think this is correct.

But let me check with another pair. Let me take $a = -2$, $b = 3$: -35/9 ≈ -3.888...

Take $a = -2. Let me take $a = -2$ and $b = 3$. Any other combination? What if I take $a = -2$ and $b = something else$?

Wait, take $a = -2$, $b = 3$: -35/9 ≈ -3.888...

Take $a = -2$, $b = 3. What if I take $a = -2$, and $b = something else$? For example, if $b = 3$, then we have the maximum of $\frac{1}{b^2}$. If I take $b$ larger, $1/b^2$ becomes smaller, so the value becomes more negative. If I take $b$ smaller, but $b$ can't be smaller than 3.

Therefore, yes, the maximum is indeed at that point.

Therefore, I think the answer is \boxed{-\dfrac{35}{9}}.

Wait, but the original problem might have wanted the answer as a positive fraction? But according to all calculations, it's negative. Let me check once again:

Original expression is $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me plug in $a = -2$, $b = 3$:

First term: $-2 + 1/3 = -5/3$

Second term: $1/3 - (-2) = 1/3 + 2 = 7/3$

Multiply: $-5/3 * 7/3 = -35/9$. Yes, negative.

Alternatively, is there a way that the expression could be positive? Let me think. Suppose $a + 1/b$ is positive and $\frac{1}{b} - a$ is positive, then their product is positive. However, as analyzed before, $a + 1/b$ is always negative. Because $a$ is as low as -6 to -2, and $1/b$ is positive but less than 1/3. So $a + 1/b$ is negative. For example, $a = -2$, $1/b = 1/3$: -2 + 1/3 = -5/3 < 0. If $a$ is more negative, say $a = -3$, $b = 3$: $-3 + 1/3 = -8/3 < 0. Even if $a$ is -2 and $b$ is very small (which it can't be, since $b$ is at least 3), even then $a + 1/b$ is -2 + 1/3 = -5/3 < 0. Therefore, the first term is always negative, the second term is always positive, so the product is always negative. Therefore, the expression is always negative, so the greatest value is the one closest to zero, which is -35/9.

Therefore, I think this is correct. So final answer is \boxed{-\dfrac{35}{9}}.

**Final Answer**
\boxed{-\dfrac{35}{9}}

To find the greatest possible value of the expression
$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)
$$
under the constraints $-6 \leq a \leq -2$ and $3 \leq b \leq 5$, we begin by simplifying the expression algebraically.

---

### **Step 1: Simplify the Expression**

Let’s denote the expression as:
$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)
$$

This is a product of the form $(x + y)(y - x)$, which simplifies to a **difference of squares**:
$$
(x + y)(y - x) = y^2 - x^2
$$

In our case, let $x = a$ and $y = \frac{1}{b}$, so we have:
$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2 = \frac{1}{b^2} - a^2
$$

So, the original expression simplifies to:
$$
\frac{1}{b^2} - a^2
$$

---

### **Step 2: Analyze the Expression**

We now want to **maximize**:
$$
\frac{1}{b^2} - a^2
$$

This expression is composed of:
- A **positive** term $\frac{1}{b^2}$, which is largest when $b$ is smallest (since the function $ \frac{1}{b^2} $ is decreasing in the interval $b \in [3, 5]$).
- A **negative** term $-a^2$, which is **closest to zero (i.e., "least negative")** when $a$ is closest to zero. In the interval $a \in [-6, -2]$, the value of $a$ closest to zero is $a = -2$.

Thus, to **maximize** the entire expression, we can set:
- $a = -2$ (to minimize $a^2$)
- $b = 3$ (to maximize $\frac{1}{b^2}$)

---

### **Step 3: Calculate the Maximum Value**

Substitute $a = -2$ and $b = 3$ into the simplified expression:
$$
\frac{1}{3^2} - (-2)^2 = \frac{1}{9} - 4 = \frac{1 - 36}{9} = \boxed{-\dfrac{35}{9}}
$$

---

### **Step 4: Confirming that This is the Maximum**

Let’s verify that replacing $a$ or $b$ with other values within the given constraints would not yield a greater result:

- For $b > 3$, $\frac{1}{b^2} < \frac{1}{9}$, which reduces the value of the expression.
- For $a < -2$, $a^2 > 4$, which increases the magnitude of the negative term, again reducing the expression value.

Thus, the maximum possible value of the expression occurs when:
- $a = -2$
- $b = 3$

---

### **Final Answer**

$$
\boxed{-\dfrac{35}{9}}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest possible value of the expression $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$ given the constraints $-6 \leq a \leq -2$ and $3 \leq b \leq 5$. Hmm, let me start by understanding what this expression is and how I can simplify or analyze it.

First, maybe I should try expanding the expression to see if it can be simplified. Let me do that:

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

This looks like a product of two terms. Let me multiply them out. Remembering the distributive property, which is like FOIL for binomials:

First terms: $a \times \frac{1}{b} = \frac{a}{b}$

Outer terms: $a \times (-a) = -a^2$

Inner terms: $\frac{1}{b} \times \frac{1}{b} = \frac{1}{b^2}$

Last terms: $\frac{1}{b} \times (-a) = -\frac{a}{b}$

Now combine all these terms:

$$
\frac{a}{b} - a^2 + \frac{1}{b^2} - \frac{a}{b}
$$

Wait, the first and the last terms are $\frac{a}{b}$ and $-\frac{a}{b}$, which cancel each other out! So that simplifies the expression to:

$$
- a^2 + \frac{1}{b^2}
$$

Oh, that's much simpler! So the original expression simplifies to $-a^2 + \frac{1}{b^2}$. Therefore, the problem reduces to maximizing $-a^2 + \frac{1}{b^2}$ given the constraints on $a$ and $b$.

Alright, so now I need to maximize $-a^2 + \frac{1}{b^2}$. Let me think about how to approach this. Since $a$ and $b$ are independent variables (their ranges don't overlap or interact except through the expression), maybe I can analyze each term separately?

But wait, actually, the expression is a combination of a function of $a$ and a function of $b$. So, since they are separate variables, maybe the maximum occurs when each term is maximized? However, I need to check if increasing one term would affect the other. Let me think.

But actually, since $-a^2$ is a function of $a$ only, and $\frac{1}{b^2}$ is a function of $b$ only, their combination is the sum of two separate functions. So, if I want to maximize the entire expression, I can maximize each term individually and then add them up. But I need to verify if that's valid here.

Wait, but suppose that maximizing $-a^2$ and maximizing $\frac{1}{b^2}$ are independent. Since $a$ and $b$ are independent variables, then yes, the maximum of the sum is the sum of the maxima. However, let me confirm this logic.

Suppose that for $-a^2$, since $a$ is between -6 and -2, then $a^2$ is between 4 and 36. Therefore, $-a^2$ is between -36 and -4. So, the maximum of $-a^2$ is -4, which occurs when $a^2$ is minimized, i.e., when $a$ is as close to zero as possible. Since $a$ is between -6 and -2, the smallest $a^2$ is when $a = -2$, because (-2)^2 = 4, which is less than (-6)^2 = 36. So maximum of $-a^2$ is -4 when a = -2.

Similarly, for $\frac{1}{b^2}$, since $b$ is between 3 and 5, $b^2$ is between 9 and 25, so $\frac{1}{b^2}$ is between 1/25 and 1/9. Therefore, maximum of $\frac{1}{b^2}$ is 1/9 when $b$ is minimized, i.e., when $b = 3$.

Therefore, if I can take $a = -2$ and $b = 3$, then the expression would be $-4 + 1/9 = -4 + 1/9 = -35/9$. But wait, is this the maximum?

But hold on, maybe there's a combination where even though $-a^2$ is not maximized, but $\frac{1}{b^2}$ is larger? Or vice versa? Wait, but since $-a^2$ is always negative, and $\frac{1}{b^2}$ is always positive. So to maximize the total expression, we need to maximize the positive term and minimize the negative term. Because $-a^2$ is negative, so to make the total expression as large as possible, we need to make it as close to zero as possible (i.e., minimize the negative part) and maximize the positive part.

So, that would mean:

Maximize $\frac{1}{b^2}$ (which is achieved when b is as small as possible, i.e., b = 3) and minimize $-a^2$ (which is equivalent to maximizing $a^2$, because $-a^2$ is negative, so minimizing it would mean making it more negative, but wait that contradicts.

Wait, hold on, maybe my earlier conclusion was wrong.

Wait, the expression is $-a^2 + \frac{1}{b^2}$. So, if I think of this as ($-a^2$) + ($\frac{1}{b^2}$). Therefore, to maximize this, I need to maximize each term. However, $-a^2$ is a negative term, so maximizing it would mean making it less negative, i.e., closer to zero. Whereas $\frac{1}{b^2}$ is positive, so maximizing it would make that term larger. Therefore, the maximum occurs when both terms are maximized. But how?

Wait, so maximizing $-a^2$ would require minimizing $a^2$, which occurs at $a = -2$ (since $a$ is negative). Then, $\frac{1}{b^2}$ is maximized when $b$ is at its minimum, which is 3. So combining these, with $a = -2$ and $b = 3$, gives us the maximum value of the expression. Which is $-(-2)^2 + (1/3)^2 = -4 + 1/9 = -35/9$. But this is negative. Wait, is this the maximum?

Wait, maybe I need to check if there are other possibilities where even though one term is smaller, the combination is larger? For example, suppose that increasing $a^2$ (so making $-a^2$ more negative) but increasing $\frac{1}{b^2}$ enough to compensate? But since $\frac{1}{b^2}$ is positive and $-a^2$ is negative, how can increasing $\frac{1}{b^2}$ compensate for a more negative term?

Wait, for example, suppose that I take $a = -6$ and $b = 5$. Then the expression would be $-(-6)^2 + (1/5)^2 = -36 + 1/25 = -35.96$, which is way smaller than -35/9 ≈ -3.888... So that's worse. If I take $a = -2$ and $b = 5$, then it's $-4 + 1/25 = -3.96$, which is still worse than -35/9 ≈ -3.888...

Wait, but if I take $a = -2$ and $b = 3$, that gives me -4 + 1/9 ≈ -3.888..., which is better. Wait, let me check if there's a way to get a higher value (i.e., less negative) by combining different a and b values?

Alternatively, maybe my initial approach is wrong? Let me check again.

Wait, the original expression, after simplifying, is $-a^2 + \frac{1}{b^2}$. So, in order to maximize this expression, since $-a^2$ is negative and $\frac{1}{b^2}$ is positive, we need to make $-a^2$ as small as possible (i.e., as close to zero as possible) and $\frac{1}{b^2}$ as large as possible. Therefore, to minimize $-a^2$, we need to maximize $a^2$. Wait, hold on, because if you have $-a^2$, then minimizing $-a^2$ would mean making $-a^2$ as small as possible, but since $a^2$ is positive, $-a^2$ is negative. Therefore, the smallest value of $-a^2$ is when $a^2$ is largest. Wait, but that's contradictory with the earlier thought. Wait, no:

Wait, suppose I have two numbers, say, -5 and -3. Which one is smaller? -5 is smaller than -3. So, if I want to minimize $-a^2$, I need to make $-a^2$ as small as possible, which would be when $a^2$ is as large as possible. Therefore, the minimum of $-a^2$ is when $a^2$ is maximum. Therefore, if I want to minimize $-a^2$, which is equivalent to maximizing $a^2$. Therefore, in order to have the expression $-a^2 + \frac{1}{b^2}$ as large as possible, since $-a^2$ is being minimized (i.e., going more negative) and $\frac{1}{b^2}$ is being maximized, but does that help?

Wait, hold on. Let me think again. If you have an expression like $X + Y$, and you want to maximize it. If X is negative and Y is positive, you need to make X as large as possible (i.e., closer to zero) and Y as large as possible. Because making X larger (less negative) and Y larger would add up to a larger total. Therefore, to maximize $X + Y$, with $X = -a^2$ and $Y = \frac{1}{b^2}$, you need to maximize $X$ (i.e., make $-a^2$ as large as possible, which is equivalent to making $a^2$ as small as possible) and maximize $Y$ (i.e., make $b$ as small as possible). Therefore, that would be the case when $a$ is as close to zero as possible (i.e., $a = -2$) and $b$ as small as possible (i.e., $b = 3$). Therefore, the maximum value is $-(-2)^2 + (1/3)^2 = -4 + 1/9 = -35/9$, which is approximately -3.888...

But wait, is there a way to get a higher value (i.e., a less negative number) by choosing different a and b?

Wait, let me check with other values. For example, suppose I choose $a = -2$ and $b = 3$, as before, giving $-4 + 1/9 = -35/9 ≈ -3.888$. What if I choose $a = -3$ and $b = 3$? Then the expression would be $-(-3)^2 + 1/9 = -9 + 1/9 = -80/9 ≈ -8.888$, which is worse. If I take $a = -2$ and $b = 5$, then it's $-4 + 1/25 = -4 + 0.04 = -3.96$, which is worse than -35/9. If I take $a = -2$ and $b = 3$, that's -35/9 ≈ -3.888...

Wait, what if I take $a = -2$ and $b = something else$? Let me think. Since $\frac{1}{b^2}$ is maximized at $b = 3$, so that's the best possible for that term. So, perhaps that is the maximum? But maybe if I take some other combination where the negative term is less negative and the positive term is not as large? Wait, but since we can't increase the positive term beyond 1/9, and we can't make the negative term less negative than -4 (since $-a^2$ is minimized at -4 when $a = -2$), then maybe that is indeed the maximum. Wait, but hold on, let me check if there's some combination where $-a^2 + \frac{1}{b^2}$ is larger than -35/9. Let's try.

Suppose we take $a = -2$ and $b = 3$, then value is -35/9 ≈ -3.888. Let me try $a = -2.5$ and $b = 3$. Then, $-a^2 = -(-2.5)^2 = -6.25$, and $\frac{1}{b^2} = 1/9$. So total is -6.25 + 1/9 ≈ -6.25 + 0.111 ≈ -6.139, which is worse. So worse.

If I take $a = -1.5$, but wait a can't be -1.5 because a is between -6 and -2. So the minimum value of a is -6 and maximum is -2. So $a$ can't be between -2 and 0. Therefore, $a$ is always negative, between -6 and -2. So the closest it can get to zero is -2. Therefore, if I take $a = -2$, that's the maximum value (closest to zero) of a. Therefore, $-a^2$ is maximized at $a = -2$ (since $-a^2$ is maximum when $a^2$ is minimum). Therefore, yes, so with that, if I also take $b = 3$, then the expression is maximized. Therefore, the maximum value is -35/9? Wait, but that seems counterintuitive because the original expression was $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$, which we simplified to $-a^2 + \frac{1}{b^2}$. But if that's the case, then maybe there's a mistake in simplification?

Wait, let me double-check the expansion. Let me redo the multiplication step by step.

Original expression:

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

Let me consider this as $(X + Y)(Y - X)$ where $X = a$ and $Y = \frac{1}{b}$. Wait, that's similar to $(Y + X)(Y - X) = Y^2 - X^2$. Wait, that's a difference of squares! So:

$$
(Y + X)(Y - X) = Y^2 - X^2
$$

Therefore, substituting back:

$$
\left(\frac{1}{b}\right)^2 - a^2 = \frac{1}{b^2} - a^2
$$

Which is the same as $-a^2 + \frac{1}{b^2}$. So that's correct. So the simplification is correct.

So then, the expression is indeed equal to $\frac{1}{b^2} - a^2$. Therefore, to maximize this, we need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Which means, maximize $\frac{1}{b^2}$ by taking the smallest possible b (which is 3) and minimize $a^2$ by taking the value of $a$ closest to 0, which is $a = -2$.

Therefore, the maximum value is $\frac{1}{3^2} - (-2)^2 = \frac{1}{9} - 4 = -\frac{35}{9}$. But this is negative. Wait, but can we get a positive value?

Wait, if the original expression was $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$, is there any chance that it can be positive? Let me check with specific values.

Take $a = -2$ and $b = 3$:

First term: $-2 + 1/3 ≈ -2 + 0.333 ≈ -1.666$

Second term: $1/3 - (-2) = 1/3 + 2 ≈ 2.333$

Multiplying them: (-1.666)(2.333) ≈ -3.888, which is the same as -35/9 ≈ -3.888...

What if I take a different combination? Let's say $a = -2$ and $b = 5$:

First term: -2 + 1/5 = -1.8

Second term: 1/5 - (-2) = 1/5 + 2 = 2.2

Product: (-1.8)(2.2) = -3.96, which is worse. So still negative.

What if I take $a = -3$ and $b = 3$:

First term: -3 + 1/3 ≈ -2.666

Second term: 1/3 - (-3) ≈ 1/3 + 3 ≈ 3.333

Product ≈ (-2.666)(3.333) ≈ -8.888...

Still negative.

Wait, so is there any possible combination where the product is positive?

Let me think. For the product $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$ to be positive, either both factors are positive or both are negative.

Let me analyze the signs of each term.

First, since $a$ is between -6 and -2, so $a$ is negative. $b$ is between 3 and 5, so $1/b$ is between 1/5 and 1/3, so positive. Therefore, $a + 1/b$ is negative plus positive. Depending on the values, this could be negative or positive?

Similarly, $\frac{1}{b} - a$ is positive minus negative, which is positive + positive, so positive. Therefore, the second term is always positive. Therefore, the product is positive times (either positive or negative). So for the entire product to be positive, the first term must also be positive. Therefore, $\left(a + \frac{1}{b}\right)$ must be positive and $\left(\frac{1}{b} - a\right)$ is always positive. Therefore, the entire expression is positive if and only if $a + \frac{1}{b} > 0$, which is equivalent to $a > -\frac{1}{b}$. Since $a$ is negative and $b$ is positive, this inequality may or may not hold.

So, if there exists values of $a$ and $b$ such that $a > -1/b$, then the product is positive. Therefore, maybe there is a maximum positive value?

Wait, this is an important point. Earlier, I thought the expression was always negative, but actually, it can be positive if $a + 1/b$ is positive. Therefore, my previous conclusion that the maximum value is -35/9 is wrong because it was assuming that the maximum occurs when the expression is negative. However, there might be positive values, which could be larger.

Therefore, I need to re-examine the problem. So the expression can be positive or negative. Therefore, the maximum value could be positive. Therefore, I need to consider both possibilities.

Let me first check whether $a + 1/b$ can be positive. That would mean $a > -1/b$. Since $a$ is between -6 and -2, and $b$ is between 3 and 5. Therefore, $-1/b$ is between -1/5 and -1/3. So, for $a > -1/b$, given that $a$ is between -6 and -2, we need to find if there exists $a$ and $b$ such that $a > -1/b$.

Let me think. Since $a$ is at least -6 and at most -2, and $-1/b$ is between -1/5 and -1/3 ≈ -0.333. Since $a$ is at least -6, but to have $a > -1/b$, which is greater than -1/3. So, if $a$ is between -2 and -1/3, then $a$ is greater than -1/b (since $b$ is at least 3, so $-1/b$ is at most -1/3). However, our constraint is $a$ between -6 and -2, so $a$ is between -6 and -2. Therefore, $a$ is always less than or equal to -2, which is less than -1/3. Wait, because -2 is approximately -2.0, which is less than -0.333. Therefore, $a$ is always less than -1/b? Wait, no. Wait, if $a$ is between -6 and -2, and $-1/b$ is between -1/5 and -1/3. For example, take $b = 5$, so $-1/b = -1/5 = -0.2$. So, if $a$ is between -6 and -2, then $a$ is less than -0.2. Therefore, $a + 1/b$ is $a + 1/b$, where $a < -0.2$ and $1/b < 0.2$. Therefore, adding them, $a + 1/b < -0.2 + 0.2 = 0$, but how much less?

Wait, let's suppose $a = -2$, $b = 5$. Then $a + 1/b = -2 + 0.2 = -1.8 < 0$. If $a = -2$, $b = 3$, $a + 1/b = -2 + 1/3 ≈ -1.666 < 0$. If $a = -2.5$, $b = 5$: $-2.5 + 0.2 = -2.3 < 0$. If $a = -1.5$, but $a$ can't be -1.5. Wait, so even if $b$ is as small as 3, then $-1/b = -1/3 ≈ -0.333$. So, if $a$ is between -6 and -2, then $a$ is always less than -0.333? Wait, no. Wait, if $a$ is between -6 and -2, then the maximum value of $a$ is -2, which is less than -1/3 ≈ -0.333? Wait no! Wait, -2 is less than -1/3? Yes, because -2 is further to the left on the number line. So $a$ is always less than -1/3, which is the upper bound of $-1/b$. Therefore, $a + 1/b < -1/b + 1/b = 0$. Therefore, $a + 1/b$ is negative for all allowed $a$ and $b$. Therefore, the first term is negative, and the second term is positive, so the entire product is negative. Therefore, the expression is always negative. Therefore, my initial conclusion that the maximum value is -35/9 is correct? But then why did I get confused earlier about positive values? Because I thought maybe there could be positive values, but actually, given the constraints, it's always negative.

Therefore, the expression is always negative, so the maximum possible value is the one closest to zero, which is achieved when the expression is least negative. Which would be when $-a^2 + \frac{1}{b^2}$ is closest to zero, i.e., when $-a^2 + \frac{1}{b^2}$ is as large as possible (since it's negative).

So, as before, to maximize $-a^2 + \frac{1}{b^2}$, which is equivalent to minimizing $a^2$ and maximizing $\frac{1}{b^2}$. Therefore, as before, take $a = -2$ (minimizing $a^2$) and $b = 3$ (maximizing $\frac{1}{b^2}$), which gives the value of $-4 + 1/9 = -35/9$. But I should check if there's another combination where the value is higher (i.e., less negative) than -35/9.

Wait, for instance, suppose I take $a = -2$ and $b = 3$: value is -35/9 ≈ -3.888...

What if I take $a = -2$ and $b$ slightly larger? Let's take $b = 4$. Then the value is $-(-2)^2 + 1/16 = -4 + 1/16 = -63/16 ≈ -3.9375$, which is worse (more negative). Similarly, if I take $b = 3.5$, then $\frac{1}{b^2} ≈ 1/(12.25) ≈ 0.0816$, so total value ≈ -4 + 0.0816 ≈ -3.9184, which is worse than -35/9 ≈ -3.888... So worse.

Alternatively, if I take $a$ slightly greater than -2 (i.e., closer to zero), but $a$ can't be greater than -2 because the upper bound is -2. So $a = -2$ is the closest to zero. What if I take $a = -2$ and $b = 3$, which gives -35/9. What if I take $a$ slightly less than -2, like $a = -2.1$ (but since $a$ can be as low as -6, but let's take a value between -6 and -2). Wait, if I take $a$ further from zero, then $-a^2$ becomes more negative. So that would make the entire expression more negative. Therefore, worse. Therefore, the maximum value is indeed achieved at $a = -2$ and $b = 3$.

Wait, but wait, there's another thought. Maybe if I take a different value of $a$ and $b$ such that $-a^2 + \frac{1}{b^2}$ is larger? For example, maybe there's a balance between decreasing $a^2$ and increasing $\frac{1}{b^2}$?

Wait, for example, suppose $a = -2$, which gives $-a^2 = -4$, and $b = 3$ gives $\frac{1}{b^2} = 1/9$. Total is -35/9 ≈ -3.888...

Alternatively, suppose I take $a = -\sqrt{\frac{1}{b^2}}$... Wait, but that's not helpful. Maybe think of the expression as a function of two variables and use calculus to find maxima?

But since this is a problem likely intended for algebra, maybe calculus is not expected. However, let's try. Maybe taking partial derivatives?

But let me think again. Since the expression is $-a^2 + \frac{1}{b^2}$, with constraints on $a$ and $b$. So to maximize this expression, we can consider that $-a^2$ is maximized when $a^2$ is minimized, which is when $a$ is closest to zero, i.e., $a = -2$. Similarly, $\frac{1}{b^2}$ is maximized when $b$ is minimized, i.e., $b = 3$. Therefore, the combination of these two gives the maximum value of the expression. Therefore, the answer is -35/9. But I need to check again if there's any other combination where the value is higher.

Wait, suppose I take $a = -2$ and $b = 3$, giving -35/9. Suppose I take $a = -2$ and $b = something else$. For example, take $b = 4$. Then the expression is -4 + 1/16 = -63/16 = -3.9375. Which is less than -35/9 (-35/9 ≈ -3.888). Similarly, if I take $a = -2$ and $b = 5$, it's -4 + 1/25 = -4 + 0.04 = -3.96, which is worse.

Alternatively, take $a$ slightly different. For example, take $a = -2$ and $b = 3$, and compare to $a = -2.5$ and $b = something$. Let me try $a = -2.5$, and find a $b$ such that maybe the expression is better. Let's compute $-a^2 + 1/b^2 = -6.25 + 1/b^2$. To make this as large as possible, we need to maximize $1/b^2$, which is when $b$ is minimized (3). Therefore, the value would be -6.25 + 1/9 ≈ -6.25 + 0.111 ≈ -6.139, which is worse than -3.888. Therefore, worse.

Alternatively, suppose I take $a = -2$ and $b = 3$, which is the best. So maybe that is indeed the maximum. Therefore, the answer is -35/9. But wait, the problem says "the greatest possible value". Since the expression is always negative, the greatest possible value is the one closest to zero, which is -35/9. But let me check one more time with different a and b values.

Suppose $a = -2$ and $b = 3$, expression is -35/9 ≈ -3.888...

What if $a = -2$ and $b = \sqrt{1/(something)}$... Wait, but $b$ is between 3 and 5, so no. Alternatively, if there is some $a$ and $b$ where the two terms balance out to give a higher value. For example, maybe $-a^2 + 1/b^2 = 0$? That would require $1/b^2 = a^2$, so $a = \pm 1/b$. However, since $a$ is negative and $1/b$ is positive, $a = -1/b$. So $a = -1/b$. Is there a $a$ and $b$ in the given ranges where this holds?

Let me solve for $a = -1/b$. Given $a$ is between -6 and -2, $b$ is between 3 and 5. Then $a = -1/b$ implies that $ -6 \leq -1/b \leq -2 $. Multiply all parts by -1 (and reverse inequalities):

$6 \geq 1/b \geq 2$, which is equivalent to $2 \leq 1/b \leq 6$. Taking reciprocals (and reversing inequalities again because all are positive):

$1/6 \leq b \leq 1/2$. But $b$ is between 3 and 5, so this is impossible. Therefore, there is no solution where $a = -1/b$ within the given ranges. Therefore, the expression cannot be zero. Therefore, the expression is always negative, so the maximum value is indeed -35/9.

But wait, hold on. Wait, maybe I can check with another approach. Let me think of the original expression before simplifying, $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me consider substituting variables. Let me denote $x = a$ and $y = \frac{1}{b}$. Then, since $a$ is between -6 and -2, $x$ is between -6 and -2. Since $b$ is between 3 and 5, $y$ is between 1/5 and 1/3.

Then the expression becomes $(x + y)(y - x) = y^2 - x^2$. So again, same result as before. So we need to maximize $y^2 - x^2$ given that $x \in [-6, -2]$ and $y \in [1/5, 1/3]$.

Therefore, since $y^2$ is maximized when $y$ is as large as possible (i.e., $y = 1/3$) and $x^2$ is minimized when $x$ is as close to zero as possible (i.e., $x = -2$), then the maximum of $y^2 - x^2$ is indeed $(1/3)^2 - (-2)^2 = 1/9 - 4 = -35/9$. So same result.

Therefore, this seems to confirm that -35/9 is the maximum value.

Wait, but the problem says "the greatest possible value". Since all values are negative, the greatest possible value is the one closest to zero, which is indeed -35/9. But maybe I need to check if there's a case where $y^2 - x^2$ is greater than -35/9? For example, take $x = -2$ and $y = 1/3$: -35/9 ≈ -3.888...

If I take $x = -2$ and $y = 1/3$: same.

If I take $x = -2$ and $y = 1/3$, then it's maximum. Any other combination?

Take $x = -2$ and $y = something less than 1/3$, then $y^2$ decreases, so the value decreases. If I take $x$ more negative, then $x^2$ increases, so $y^2 - x^2$ decreases. Therefore, the maximum occurs at $x = -2$, $y = 1/3$. Therefore, the answer is -35/9.

But wait, the problem says "the greatest possible value", and asks to express it as a common fraction. -35/9 is a common fraction. However, is there a possibility that I made a mistake in the sign?

Wait, let me check once again. Original expression:

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

Expanding gives $\frac{1}{b^2} - a^2$, which is the same as $-a^2 + \frac{1}{b^2}$. So yes.

But let me compute with actual numbers. Take $a = -2$, $b = 3$:

First factor: $-2 + 1/3 = -5/3$

Second factor: $1/3 - (-2) = 1/3 + 2 = 7/3$

Product: (-5/3)(7/3) = -35/9. Correct. Therefore, the value is indeed -35/9.

But is there a possibility that the problem expects a positive answer? Maybe I misread the problem? Let me check the original problem again:

"If $-6 \leq a \leq -2$ and $3 \leq b \leq 5$, what is the greatest possible value of $\left(a+\frac{1}{b}\right)\left(\frac{1}{b}-a\right)$? Express your answer as a common fraction."

No, it says "greatest possible value", which is the maximum value. Since all values are negative, the maximum is the least negative one, which is -35/9. Therefore, unless there's a miscalculation, that should be the answer.

Wait, but maybe I need to check if there's another way to interpret the original expression. Maybe there was a typo in my expansion?

Wait, original expression: $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me multiply it again:

First, multiply $a$ by $\frac{1}{b}$: $a/b$

Then, $a$ times $-a$: $-a^2$

Then, $\frac{1}{b}$ times $\frac{1}{b}$: $1/b^2$

Then, $\frac{1}{b}$ times $-a$: $-a/b$

So altogether:

$a/b - a^2 + 1/b^2 - a/b$

Yes, the $a/b$ and $-a/b$ cancel, leaving $-a^2 + 1/b^2$, which is correct. So that's right.

Another way: note that $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2$, which is difference of squares. So that's correct.

Therefore, I think the answer is indeed -35/9.

But just to be thorough, let me check if there's any other combination where $-a^2 + \frac{1}{b^2}$ is larger. For example, take $a = -2$ and $b = 3$, as before: -4 + 1/9 = -35/9 ≈ -3.888...

What if $a = -2$ and $b = 3. Let me take $a$ slightly different. Suppose $a = -2$ and $b$ is variable. Since $-a^2$ is fixed at -4, and $\frac{1}{b^2}$ is maximum at 1/9, so that's the best possible for that combination. If I take $a$ varying, but keeping $a$ at -2 is optimal. So no.

Alternatively, maybe take $a$ slightly different and $b$ slightly different such that the decrease in $-a^2$ is offset by an increase in $\frac{1}{b^2}$. For example, suppose I take $a = -2. Let me pick a $a$ slightly larger than -2 (i.e., closer to zero) but since the maximum $a$ is -2, I can't go any closer. So no. If I take $a$ slightly less than -2 (i.e., further from zero), which would decrease $-a^2$, making it more negative. Therefore, worse.

Alternatively, if I take $b$ slightly smaller than 3, but since $b$ can't be smaller than 3, as given $3 \leq b \leq 5$, so no.

Therefore, there's no way to get a higher value than -35/9.

Wait, but just to check, maybe if $a$ and $b$ are chosen such that $-a^2 + \frac{1}{b^2}$ is larger? Let me think of extremes. For example, when $a = -6$, $b = 5$: $-36 + 1/25 = -35.96$, which is worse. When $a = -2$, $b = 5$: -4 + 1/25 ≈ -3.96. When $a = -2$, $b = 3$: -35/9 ≈ -3.888... So indeed, the latter is higher.

Alternatively, take $a = -2$, $b = 3$: this is the best. Therefore, I think I can be confident that -35/9 is the correct answer.

But just to check with another method. Let me consider that the expression is $-a^2 + \frac{1}{b^2}$. Since both $a$ and $b$ are independent variables, the maximum occurs at the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$.

Since $a$ is between -6 and -2, $a^2$ is between 4 and 36. Therefore, the minimum $a^2$ is 4, achieved at $a = -2$. The maximum $\frac{1}{b^2}$ is 1/9, achieved at $b = 3$. Therefore, maximum value is 1/9 - 4 = -35/9. So same answer.

Therefore, I think this confirms that the answer is indeed -35/9.

But the problem says "the greatest possible value", and the answer is negative. However, maybe I made a mistake in simplifying the expression? Wait, let me check once again with specific numbers. Let me take $a = -2$, $b = 3$. Then the original expression:

First factor: $-2 + 1/3 = -5/3$

Second factor: $1/3 - (-2) = 1/3 + 2 = 7/3$

Product: (-5/3)(7/3) = -35/9. Correct.

If I take $a = -2$, $b = 3$, the value is -35/9. If I take $a = -2$, $b = 3$, is that within the given ranges? Yes, since $-6 \leq -2 \leq -2$ and $3 \leq 3 \leq 5$. So yes.

Therefore, the answer is -35/9. But since the problem says "greatest possible value", and -35/9 is indeed the maximum (since all other combinations give smaller values), then I think that is the answer.

However, just to be 100% sure, let me check another combination where maybe the product is higher. Let me take $a = -2$, $b = 3$: -35/9. What if I take $a = -2$, $b$ approaching 3 from above? Then $1/b^2$ approaches 1/9 from below, so the expression approaches -35/9 from below. So worse. If I take $a$ approaching -2 from below (i.e., more negative), then $-a^2$ becomes more negative, so the expression becomes worse. Therefore, no.

Alternatively, take $a = -2$ and $b = 3$: maximum.

Therefore, conclusion: the greatest possible value is -35/9, which is a common fraction.

But wait, the problem might expect a positive answer? Let me check once again for any possible mistake. Wait, could there be a miscalculation in the expansion?

Original expression:

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

Let me compute this expression with $a = -2$, $b = 3$:

First factor: $-2 + 1/3 = -5/3$

Second factor: $1/3 - (-2) = 1/3 + 2 = 7/3$

Product: $(-5/3)(7/3) = -35/9$. Correct.

Alternatively, if I consider $\frac{1}{b^2} - a^2$ with the same values: $1/9 - 4 = -35/9$. Correct.

Therefore, I think the answer is correct. Therefore, the greatest possible value is $\boxed{-\dfrac{35}{9}}$.

Wait, but the problem says "greatest possible value", and if all values are negative, then the greatest is the one closest to zero. However, sometimes problems might have a trick where the expression can be positive. But according to all the above, it can't. Therefore, unless there's a mistake in my analysis, I think this is correct.

But let me check with another pair. Let me take $a = -2$, $b = 3$: -35/9 ≈ -3.888...

Take $a = -2. Let me take $a = -2$ and $b = 3$. Any other combination? What if I take $a = -2$ and $b = something else$?

Wait, take $a = -2$, $b = 3$: -35/9 ≈ -3.888...

Take $a = -2$, $b = 3. What if I take $a = -2$, and $b = something else$? For example, if $b = 3$, then we have the maximum of $\frac{1}{b^2}$. If I take $b$ larger, $1/b^2$ becomes smaller, so the value becomes more negative. If I take $b$ smaller, but $b$ can't be smaller than 3.

Therefore, yes, the maximum is indeed at that point.

Therefore, I think the answer is \boxed{-\dfrac{35}{9}}.

Wait, but the original problem might have wanted the answer as a positive fraction? But according to all calculations, it's negative. Let me check once again:

Original expression is $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me plug in $a = -2$, $b = 3$:

First term: $-2 + 1/3 = -5/3$

Second term: $1/3 - (-2) = 1/3 + 2 = 7/3$

Multiply: $-5/3 * 7/3 = -35/9$. Yes, negative.

Alternatively, is there a way that the expression could be positive? Let me think. Suppose $a + 1/b$ is positive and $\frac{1}{b} - a$ is positive, then their product is positive. However, as analyzed before, $a + 1/b$ is always negative. Because $a$ is as low as -6 to -2, and $1/b$ is positive but less than 1/3. So $a + 1/b$ is negative. For example, $a = -2$, $1/b = 1/3$: -2 + 1/3 = -5/3 < 0. If $a$ is more negative, say $a = -3$, $b = 3$: $-3 + 1/3 = -8/3 < 0. Even if $a$ is -2 and $b$ is very small (which it can't be, since $b$ is at least 3), even then $a + 1/b$ is -2 + 1/3 = -5/3 < 0. Therefore, the first term is always negative, the second term is always positive, so the product is always negative. Therefore, the expression is always negative, so the greatest value is the one closest to zero, which is -35/9.

Therefore, I think this is correct. So final answer is \boxed{-\dfrac{35}{9}}.

**Final Answer**
\boxed{-\dfrac{35}{9}}

To find the greatest possible value of the expression
$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)
$$
under the constraints $-6 \leq a \leq -2$ and $3 \leq b \leq 5$, we begin by simplifying the expression algebraically.

---

### **Step 1: Simplify the Expression**

Let’s denote the expression as:
$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)
$$

This is a product of the form $(x + y)(y - x)$, which simplifies to a **difference of squares**:
$$
(x + y)(y - x) = y^2 - x^2
$$

In our case, let $x = a$ and $y = \frac{1}{b}$, so we have:
$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2 = \frac{1}{b^2} - a^2
$$

So, the original expression simplifies to:
$$
\frac{1}{b^2} - a^2
$$

---

### **Step 2: Analyze the Expression**

We now want to **maximize**:
$$
\frac{1}{b^2} - a^2
$$

This expression is composed of:
- A **positive** term $\frac{1}{b^2}$, which is largest when $b$ is smallest (since the function $ \frac{1}{b^2} $ is decreasing in the interval $b \in [3, 5]$).
- A **negative** term $-a^2$, which is **closest to zero (i.e., "least negative")** when $a$ is closest to zero. In the interval $a \in [-6, -2]$, the value of $a$ closest to zero is $a = -2$.

Thus, to **maximize** the entire expression, we can set:
- $a = -2$ (to minimize $a^2$)
- $b = 3$ (to maximize $\frac{1}{b^2}$)

---

### **Step 3: Calculate the Maximum Value**

Substitute $a = -2$ and $b = 3$ into the simplified expression:
$$
\frac{1}{3^2} - (-2)^2 = \frac{1}{9} - 4 = \frac{1 - 36}{9} = \boxed{-\dfrac{35}{9}}
$$

---

### **Step 4: Confirming that This is the Maximum**

Let’s verify that replacing $a$ or $b$ with other values within the given constraints would not yield a greater result:

- For $b > 3$, $\frac{1}{b^2} < \frac{1}{9}$, which reduces the value of the expression.
- For $a < -2$, $a^2 > 4$, which increases the magnitude of the negative term, again reducing the expression value.

Thus, the maximum possible value of the expression occurs when:
- $a = -2$
- $b = 3$

---

### **Final Answer**

$$
\boxed{-\dfrac{35}{9}}
$$

Solution: The given expression expands to $\frac{1}{b^2} - a^2$.  Thus we want $b$ to have the least possible magnitude and $a$ to also have the least possible magnitude.  Our maximum value is therefore $\frac{1}{3^2} - (-2)^2 = \boxed{-\frac{35}{9}}$.