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 \( f(\theta) \) given by:

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

We are required to find the value of \( f\left(\frac{\pi}{18}\right) + f\left(\frac{7\pi}{18}\right) \).

### Step 1: Calculate \( f\left(\frac{\pi}{18}\right) \)

Substituting \( \theta = \frac{\pi}{18} \):

\[ f\left(\frac{\pi}{18}\right) = \sin^3\left(\frac{\pi}{18}\right) + \sin^3\left(\frac{\pi}{18} + \frac{2\pi}{3}\right) + \sin^3\left(\frac{\pi}{18} + \frac{4\pi}{3}\right) \]

Calculating the angles:

1. \( \frac{\pi}{18} + \frac{2\pi}{3} = \frac{\pi}{18} + \frac{12\pi}{18} = \frac{13\pi}{18} \)
2. \( \frac{\pi}{18} + \frac{4\pi}{3} = \frac{\pi}{18} + \frac{24\pi}{18} = \frac{25\pi}{18} \)

So we have:

\[ f\left(\frac{\pi}{18}\right) = \sin^3\left(\frac{\pi}{18}\right) + \sin^3\left(\frac{13\pi}{18}\right) + \sin^3\left(\frac{25\pi}{18}\right) \]

### Step 2: Simplify \( \sin^3\left(\frac{13\pi}{18}\right) \) and \( \sin^3\left(\frac{25\pi}{18}\right) \)

Using the property \( \sin\left(\pi - x\right) = \sin x \):

\[ \sin\left(\frac{13\pi}{18}\right) = \sin\left(\pi - \frac{5\pi}{18}\right) = \sin\left(\frac{5\pi}{18}\right) \]

And for \( \sin\left(\frac{25\pi}{18}\right) \):

\[ \sin\left(\frac{25\pi}{18}\right) = \sin\left(2\pi - \frac{7\pi}{18}\right) = -\sin\left(\frac{7\pi}{18}\right) \]

Thus,

\[ f\left(\frac{\pi}{18}\right) = \sin^3\left(\frac{\pi}{18}\right) + \sin^3\left(\frac{5\pi}{18}\right) - \sin^3\left(\frac{7\pi}{18}\right) \]

### Step 3: Calculate \( f\left(\frac{7\pi}{18}\right) \)

Now substituting \( \theta = \frac{7\pi}{18} \):

\[ f\left(\frac{7\pi}{18}\right) = \sin^3\left(\frac{7\pi}{18}\right) + \sin^3\left(\frac{7\pi}{18} + \frac{2\pi}{3}\right) + \sin^3\left(\frac{7\pi}{18} + \frac{4\pi}{3}\right) \]

Calculating the angles:

1. \( \frac{7\pi}{18} + \frac{2\pi}{3} = \frac{7\pi}{18} + \frac{12\pi}{18} = \frac{19\pi}{18} \)
2. \( \frac{7\pi}{18} + \frac{4\pi}{3} = \frac{7\pi}{18} + \frac{24\pi}{18} = \frac{31\pi}{18} \)

Thus,

\[ f\left(\frac{7\pi}{18}\right) = \sin^3\left(\frac{7\pi}{18}\right) + \sin^3\left(\frac{19\pi}{18}\right) + \sin^3\left(\frac{31\pi}{18}\right) \]

Using the properties of sine:

\[ \sin\left(\frac{19\pi}{18}\right) = -\sin\left(\frac{\pi}{18}\right) \quad \text{and} \quad \sin\left(\frac{31\pi}{18}\right) = -\sin\left(\frac{7\pi}{18}\right) \]

So we have:

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

### Step 4: Combine the results

Now we can add the two results:

\[ f\left(\frac{\pi}{18}\right) + f\left(\frac{7\pi}{18}\right) = \left(\sin^3\left(\frac{\pi}{18}\right) + \sin^3\left(\frac{5\pi}{18}\right) - \sin^3\left(\frac{7\pi}{18}\right)\right) + \left(\sin^3\left(\frac{7\pi}{18}\right) - \sin^3\left(\frac{\pi}{18}\right) - \sin^3\left(\frac{7\pi}{18}\right)\right) \]

This simplifies to:

\[ f\left(\frac{\pi}{18}\right) + f\left(\frac{7\pi}{18}\right) = \sin^3\left(\frac{5\pi}{18}\right) \]

### Final Result

Thus, the value of \( f\left(\frac{\pi}{18}\right) + f\left(\frac{7\pi}{18}\right) \) is:

\[ \sin^3\left(\frac{5\pi}{18}\right) \]