IF `-1lea le2and-3leb le2`, What is the greatest possible value of (a+b)(b-a)?

Answer

To solve the problem, we need to find the greatest possible value of the expression \((a + b)(b - a)\) given the constraints \( -1 \leq a \leq 2 \) and \( -3 \leq b \leq 2 \).

### Step-by-Step Solution:

1. **Rewrite the Expression**:
The expression \((a + b)(b - a)\) can be rewritten using the difference of squares:
\[ (a + b)(b - a) = b^2 - a^2 \]

2. **Identify the Ranges**:
We know from the problem that:
\[ -1 \leq a \leq 2 \quad \text{and} \quad -3 \leq b \leq 2 \]

3. **Determine Maximum \(b^2\)**:
To maximize \(b^2\), we look at the endpoints of the range for \(b\):
- When \(b = -3\), \(b^2 = (-3)^2 = 9\)
- When \(b = 2\), \(b^2 = 2^2 = 4\)

The maximum value of \(b^2\) is therefore \(9\) when \(b = -3\).

4. **Determine Minimum \(a^2\)**:
To minimize \(a^2\), we also look at the endpoints of the range for \(a\):
- When \(a = -1\), \(a^2 = (-1)^2 = 1\)
- When \(a = 0\), \(a^2 = 0^2 = 0\)
- When \(a = 2\), \(a^2 = 2^2 = 4\)

The minimum value of \(a^2\) is \(0\) when \(a = 0\).

5. **Calculate \(b^2 - a^2\)**:
Now that we have the maximum \(b^2\) and the minimum \(a^2\):
\[ b^2 - a^2 = 9 - 0 = 9 \]

6. **Conclusion**:
The greatest possible value of the expression \((a + b)(b - a)\) is:
\[ \boxed{9} \]