`tan^6pi/9-33tan^4pi/9+27tan^2pi/9` is equal to (a) 0 (b) `sqrt(3)` (c) 3 (d) 9

Options

A. 0
B. `sqrt(3)`
C. 3
D. 9

Explanation

To solve the expression \( \tan^6\left(\frac{\pi}{9}\right) - 33\tan^4\left(\frac{\pi}{9}\right) + 27\tan^2\left(\frac{\pi}{9}\right) \), we can use the identity for \( \tan(3\theta) \).

### Step 1: Use the identity for \( \tan(3\theta) \)

The identity states:
\[ \tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)} \]

Let \( \theta = \frac{\pi}{9} \). Therefore, \( 3\theta = \frac{\pi}{3} \).

### Step 2: Substitute \( \theta \) into the identity

We know that \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \). Thus, we can write:
\[ \sqrt{3} = \frac{3\tan\left(\frac{\pi}{9}\right) - \tan^3\left(\frac{\pi}{9}\right)}{1 - 3\tan^2\left(\frac{\pi}{9}\right)} \]

### Step 3: Cross-multiply

Cross-multiplying gives:
\[ \sqrt{3}(1 - 3\tan^2\left(\frac{\pi}{9}\right)) = 3\tan\left(\frac{\pi}{9}\right) - \tan^3\left(\frac{\pi}{9}\right) \]

### Step 4: Rearrange the equation

Rearranging the equation gives:
\[ \sqrt{3} - 3\sqrt{3}\tan^2\left(\frac{\pi}{9}\right) = 3\tan\left(\frac{\pi}{9}\right) - \tan^3\left(\frac{\pi}{9}\right) \]

### Step 5: Square both sides

Squaring both sides results in:
\[ 3 = (3\tan\left(\frac{\pi}{9}\right) - \tan^3\left(\frac{\pi}{9}\right))^2 + 9\tan^2\left(\frac{\pi}{9}\right)(\sqrt{3})^2 \]
This simplifies to:
\[ 3 = (3\tan\left(\frac{\pi}{9}\right) - \tan^3\left(\frac{\pi}{9}\right))^2 + 27\tan^2\left(\frac{\pi}{9}\right) \]

### Step 6: Expand the squared term

Expanding the squared term gives:
\[ 3 = 9\tan^2\left(\frac{\pi}{9}\right) - 6\tan^4\left(\frac{\pi}{9}\right) + \tan^6\left(\frac{\pi}{9}\right) + 27\tan^2\left(\frac{\pi}{9}\right) \]
Combining like terms results in:
\[ 3 = \tan^6\left(\frac{\pi}{9}\right) + 36\tan^2\left(\frac{\pi}{9}\right) - 6\tan^4\left(\frac{\pi}{9}\right) \]

### Step 7: Rearranging the equation

Rearranging gives:
\[ \tan^6\left(\frac{\pi}{9}\right) - 6\tan^4\left(\frac{\pi}{9}\right) + 36\tan^2\left(\frac{\pi}{9}\right) - 3 = 0 \]

### Step 8: Substitute \( x = \tan^2\left(\frac{\pi}{9}\right) \)

Let \( x = \tan^2\left(\frac{\pi}{9}\right) \). The equation becomes:
\[ x^3 - 6x^2 + 36x - 3 = 0 \]

### Step 9: Factor the polynomial

Using synthetic division or other methods, we can find that this polynomial has a root at \( x = 3 \).

### Step 10: Conclusion

Thus, substituting back gives:
\[ \tan^2\left(\frac{\pi}{9}\right) = 3 \]
Therefore, the original expression evaluates to:
\[ \tan^6\left(\frac{\pi}{9}\right) - 33\tan^4\left(\frac{\pi}{9}\right) + 27\tan^2\left(\frac{\pi}{9}\right) = 3 \]

The final answer is:
\[ \boxed{3} \]