Value of `tan ^(6) 40^(@) -33 tan ^(4) 40^(@)+ 27 tan ^(2) 40^(@)` is `"_________"`

Answer

To solve the expression \( \tan^6 40^\circ - 33 \tan^4 40^\circ + 27 \tan^2 40^\circ \), we can use the formula for \( \tan 3\theta \).

### Step-by-Step Solution:

1. **Substitute \( \theta \) with \( 40^\circ \)**:
We will use the formula for \( \tan 3\theta \):
\[ \tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} \]
Setting \( \theta = 40^\circ \), we have:
\[ \tan 120^\circ = \frac{3\tan 40^\circ - \tan^3 40^\circ}{1 - 3\tan^2 40^\circ} \]

2. **Evaluate \( \tan 120^\circ \)**:
We know that:
\[ \tan 120^\circ = -\sqrt{3} \]
Therefore, we can write:
\[ -\sqrt{3} = \frac{3\tan 40^\circ - \tan^3 40^\circ}{1 - 3\tan^2 40^\circ} \]

3. **Cross-multiply**:
Multiply both sides by \( 1 - 3\tan^2 40^\circ \):
\[ -\sqrt{3}(1 - 3\tan^2 40^\circ) = 3\tan 40^\circ - \tan^3 40^\circ \]

4. **Expand the left side**:
\[ -\sqrt{3} + 3\sqrt{3}\tan^2 40^\circ = 3\tan 40^\circ - \tan^3 40^\circ \]

5. **Rearranging the equation**:
Rearranging gives:
\[ \tan^3 40^\circ - 3\tan 40^\circ + 3\sqrt{3}\tan^2 40^\circ + \sqrt{3} = 0 \]

6. **Square both sides**:
To eliminate the square root, we can square both sides:
\[ 3 = \tan^6 40^\circ - 33\tan^4 40^\circ + 27\tan^2 40^\circ \]

7. **Final Result**:
Thus, we find that:
\[ \tan^6 40^\circ - 33\tan^4 40^\circ + 27\tan^2 40^\circ = 3 \]

### Conclusion:
The value of \( \tan^6 40^\circ - 33 \tan^4 40^\circ + 27 \tan^2 40^\circ \) is \( \boxed{3} \).