If `x = tan'(pi)/(18) ` then `3x^(6) - 27x^(4) + 33x^(2)` is equal to -

Options

A. 1
B. 2
C. `3sqrt(3)`
D. `1//3`

Answer

To solve the problem, we need to evaluate the expression \(3x^6 - 27x^4 + 33x^2\) given that \(x = \tan\left(\frac{\pi}{18}\right)\).

### Step-by-Step Solution:

1. **Identify the given value of x**:
\[ x = \tan\left(\frac{\pi}{18}\right) \]

2. **Use the triple angle formula for tangent**:
The formula for \(\tan(3\theta)\) is:
\[ \tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)} \]
Here, let \(\theta = \frac{\pi}{18}\). Thus, we have:
\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]

3. **Substituting into the formula**:
Substitute \(x\) into the formula:
\[ \tan\left(\frac{\pi}{6}\right) = \frac{3x - x^3}{1 - 3x^2} \]
This gives us:
\[ \frac{1}{\sqrt{3}} = \frac{3x - x^3}{1 - 3x^2} \]

4. **Cross-multiply**:
Cross-multiplying yields:
\[ 1 - 3x^2 = \sqrt{3}(3x - x^3) \]
Expanding the right side:
\[ 1 - 3x^2 = 3\sqrt{3}x - \sqrt{3}x^3 \]

5. **Rearranging the equation**:
Rearranging gives:
\[ \sqrt{3}x^3 - 3x^2 - 3\sqrt{3}x + 1 = 0 \]

6. **Square both sides**:
To eliminate the square root, we can square both sides:
\[ \left(1 - 3x^2\right)^2 = 3(3x - x^3)^2 \]

7. **Expanding both sides**:
The left side becomes:
\[ 1 - 6x^2 + 9x^4 \]
The right side expands to:
\[ 3(9x^2 - 6x^4 + x^6) = 27x^2 - 18x^4 + 3x^6 \]

8. **Setting the equation**:
Now we have:
\[ 1 - 6x^2 + 9x^4 = 27x^2 - 18x^4 + 3x^6 \]

9. **Rearranging terms**:
Rearranging gives:
\[ 3x^6 - 27x^4 + 33x^2 - 1 = 0 \]

10. **Conclusion**:
Thus, we find that:
\[ 3x^6 - 27x^4 + 33x^2 = 1 \]

### Final Answer:
\[ 3x^6 - 27x^4 + 33x^2 = 1 \]