If `f( theta ) = sin ^(3)theta + sin ^(3)( theta + (2pi)/(3)) + sin ^(3)( theta + (4pi)/(3))` then the value of `f((pi)/(18)) + f((7pi)/(18)) ` is ___________.

Explanation

To solve the problem, we need to evaluate the function \( f(\theta) = \sin^3 \theta + \sin^3 \left( \theta + \frac{2\pi}{3} \right) + \sin^3 \left( \theta + \frac{4\pi}{3} \right) \) and find the value of \( f\left( \frac{\pi}{18} \right) + f\left( \frac{7\pi}{18} \right) \).

### Step 1: Simplify the function \( f(\theta) \)

We start with the expression for \( f(\theta) \):

\[ f(\theta) = \sin^3 \theta + \sin^3 \left( \theta + \frac{2\pi}{3} \right) + \sin^3 \left( \theta + \frac{4\pi}{3} \right) \]

Using the identity for the sine function, we know that:

\[ \sin\left(\theta + \frac{2\pi}{3}\right) = -\frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta \]
\[ \sin\left(\theta + \frac{4\pi}{3}\right) = -\frac{1}{2} \sin \theta - \frac{\sqrt{3}}{2} \cos \theta \]

### Step 2: Use the identity for \( \sin^3 x \)

We can use the identity:

\[ \sin^3 x = \frac{3 \sin x - \sin(3x)}{4} \]

Applying this identity to each term in \( f(\theta) \):

\[ f(\theta) = \frac{3 \sin \theta - \sin(3\theta)}{4} + \frac{3 \sin\left(\theta + \frac{2\pi}{3}\right) - \sin\left(3\left(\theta + \frac{2\pi}{3}\right)\right)}{4} + \frac{3 \sin\left(\theta + \frac{4\pi}{3}\right) - \sin\left(3\left(\theta + \frac{4\pi}{3}\right)\right)}{4} \]

### Step 3: Combine the terms

Combining the terms, we find that:

\[ f(\theta) = \frac{1}{4} \left( 3 \left( \sin \theta + \sin\left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta + \frac{4\pi}{3}\right) \right) - \left( \sin(3\theta) + \sin\left(3\left(\theta + \frac{2\pi}{3}\right)\right) + \sin\left(3\left(\theta + \frac{4\pi}{3}\right)\right) \right) \right) \]

### Step 4: Evaluate \( f\left( \frac{\pi}{18} \right) \) and \( f\left( \frac{7\pi}{18} \right) \)

Now we calculate \( f\left( \frac{\pi}{18} \right) \):

\[ f\left( \frac{\pi}{18} \right) = \frac{3}{4} \sin\left( \frac{\pi}{18} \right) - \frac{3}{4} \sin\left( \frac{\pi}{6} \right) = \frac{3}{4} \left( \sin\left( \frac{\pi}{18} \right) - \frac{1}{2} \right) \]

Next, we calculate \( f\left( \frac{7\pi}{18} \right) \):

\[ f\left( \frac{7\pi}{18} \right) = \frac{3}{4} \sin\left( \frac{7\pi}{18} \right) - \frac{3}{4} \sin\left( \frac{7\pi}{6} \right) \]

Since \( \sin\left( \frac{7\pi}{6} \right) = -\frac{1}{2} \):

\[ f\left( \frac{7\pi}{18} \right) = \frac{3}{4} \sin\left( \frac{7\pi}{18} \right) + \frac{3}{8} \]

### Step 5: Add the two results

Now we add the two results:

\[ f\left( \frac{\pi}{18} \right) + f\left( \frac{7\pi}{18} \right) = \frac{3}{4} \left( \sin\left( \frac{\pi}{18} \right) + \sin\left( \frac{7\pi}{18} \right) \right) \]

Using the sine addition formula, we find that:

\[ \sin\left( \frac{\pi}{18} \right) + \sin\left( \frac{7\pi}{18} \right) = 0 \]

Thus, we conclude:

\[ f\left( \frac{\pi}{18} \right) + f\left( \frac{7\pi}{18} \right) = 0 \]

### Final Answer

The value of \( f\left( \frac{\pi}{18} \right) + f\left( \frac{7\pi}{18} \right) \) is \( \boxed{0} \).