If `sin^(3)theta+sin^(3)(theta+(2pi)/(3))+sin^(3)(theta+(4pi)/(3))=a sinb theta`. Find the value of `|(b)/(a)|`.

Answer

To solve the equation \( \sin^3 \theta + \sin^3 \left( \theta + \frac{2\pi}{3} \right) + \sin^3 \left( \theta + \frac{4\pi}{3} \right) = a \sin b \theta \), we will use the identity for \( \sin^3 \theta \) and properties of sine functions.

### Step-by-Step Solution:

1. **Use the identity for \( \sin^3 \theta \)**:
\[ \sin^3 \theta = \frac{3 \sin \theta - \sin 3\theta}{4} \]
Therefore, we can rewrite each term:
\[ \sin^3 \theta = \frac{3 \sin \theta - \sin 3\theta}{4} \]
\[ \sin^3 \left( \theta + \frac{2\pi}{3} \right) = \frac{3 \sin \left( \theta + \frac{2\pi}{3} \right) - \sin \left( 3\theta + 2\pi \right)}{4} \]
\[ \sin^3 \left( \theta + \frac{4\pi}{3} \right) = \frac{3 \sin \left( \theta + \frac{4\pi}{3} \right) - \sin \left( 3\theta + 4\pi \right)}{4} \]

2. **Substituting the identities into the equation**:
\[ \sin^3 \theta + \sin^3 \left( \theta + \frac{2\pi}{3} \right) + \sin^3 \left( \theta + \frac{4\pi}{3} \right) = \frac{1}{4} \left( 3 \sin \theta - \sin 3\theta + 3 \sin \left( \theta + \frac{2\pi}{3} \right) - \sin \left( 3\theta + 2\pi \right) + 3 \sin \left( \theta + \frac{4\pi}{3} \right) - \sin \left( 3\theta + 4\pi \right) \right) \]

3. **Simplifying the sine terms**:
Since \( \sin(3\theta + 2\pi) = \sin(3\theta) \) and \( \sin(3\theta + 4\pi) = \sin(3\theta) \), we can simplify:
\[ = \frac{1}{4} \left( 3 \sin \theta + 3 \sin \left( \theta + \frac{2\pi}{3} \right) + 3 \sin \left( \theta + \frac{4\pi}{3} \right) - 2 \sin 3\theta \right) \]

4. **Using the sine addition formulas**:
The sum \( \sin \left( \theta + \frac{2\pi}{3} \right) + \sin \left( \theta + \frac{4\pi}{3} \right) \) can be simplified:
\[ \sin \left( \theta + \frac{2\pi}{3} \right) + \sin \left( \theta + \frac{4\pi}{3} \right) = \sin \theta \left( \cos \frac{2\pi}{3} + \cos \frac{4\pi}{3} \right) + \cos \theta \left( \sin \frac{2\pi}{3} + \sin \frac{4\pi}{3} \right) \]
Noting that \( \cos \frac{2\pi}{3} = -\frac{1}{2} \) and \( \cos \frac{4\pi}{3} = -\frac{1}{2} \) gives:
\[ = -\sin \theta + 0 = -\sin \theta \]

5. **Final simplification**:
Thus, we have:
\[ \sin^3 \theta + \sin^3 \left( \theta + \frac{2\pi}{3} \right) + \sin^3 \left( \theta + \frac{4\pi}{3} \right) = \frac{1}{4} \left( 3 \sin \theta - 2 \sin 3\theta \right) \]
This can be rewritten as:
\[ = \frac{3}{4} \sin \theta - \frac{1}{2} \sin 3\theta \]

6. **Setting equal to \( a \sin b \theta \)**:
We can express this as:
\[ = -\frac{3}{4} \sin 3\theta \]
Thus, comparing with \( a \sin b \theta \):
\[ a = -\frac{3}{4}, \quad b = 3 \]

7. **Finding \( \left| \frac{b}{a} \right| \)**:
\[ \left| \frac{b}{a} \right| = \left| \frac{3}{-\frac{3}{4}} \right| = \left| -4 \right| = 4 \]

### Final Answer:
\[ \boxed{4} \]