Let `z=9+b i ,w h e r eb` is nonzero real and `i^2=-1.` If the imaginary part of `z^2a n dz^3` are equal, then `b//3` is ______.

Explanation

To solve the problem, we need to find the value of \( \frac{b}{3} \) given that the imaginary parts of \( z^2 \) and \( z^3 \) are equal, where \( z = 9 + bi \) and \( b \) is a non-zero real number.

### Step-by-Step Solution:

1. **Define \( z \)**:
\[ z = 9 + bi \]

2. **Calculate \( z^2 \)**:
\[ z^2 = (9 + bi)^2 = 9^2 + 2 \cdot 9 \cdot (bi) + (bi)^2 \]
\[ = 81 + 18bi + b^2(-1) = 81 - b^2 + 18bi \]
The imaginary part of \( z^2 \) is \( 18b \).

3. **Calculate \( z^3 \)**:
\[ z^3 = (9 + bi)(z^2) = (9 + bi)(81 - b^2 + 18bi) \]
Expanding this:
\[ = 9(81 - b^2) + 9(18bi) + bi(81 - b^2) + bi(18bi) \]
\[ = 729 - 9b^2 + 162bi + 81bi - b^3(-1) \]
\[ = 729 - 9b^2 + 162bi + 81bi + b^3 \]
\[ = 729 - 9b^2 + b^3 + (162b + 81b)i \]
\[ = 729 - 9b^2 + b^3 + 243bi \]
The imaginary part of \( z^3 \) is \( 243b \).

4. **Set the imaginary parts equal**:
\[ 18b = 243b - b^3 \]
Rearranging gives:
\[ b^3 - 225b = 0 \]
Factoring out \( b \):
\[ b(b^2 - 225) = 0 \]
This gives us:
\[ b = 0 \quad \text{or} \quad b^2 - 225 = 0 \]

5. **Solve for \( b \)**:
\[ b^2 = 225 \implies b = \pm 15 \]
Since \( b \) is non-zero, we take \( b = 15 \) or \( b = -15 \).

6. **Calculate \( \frac{b}{3} \)**:
\[ \frac{b}{3} = \frac{15}{3} = 5 \quad \text{or} \quad \frac{-15}{3} = -5 \]

Thus, the final answer is:
\[ \frac{b}{3} = \pm 5 \]