The value of
`sin^(6)((pi)/(49))+cos^(6)((pi)/(49))-1+3sin^(2)((pi)/(49))cos^(2)((pi)/(49))` is equal to

Options

A. `tan^(6)((pi)/(49))`
B. `cot^(6)((pi)/(49))`
C. 1
D. 0

Explanation

To solve the expression

\[ \sin^6\left(\frac{\pi}{49}\right) + \cos^6\left(\frac{\pi}{49}\right) - 1 + 3\sin^2\left(\frac{\pi}{49}\right)\cos^2\left(\frac{\pi}{49}\right), \]

we can use the identity for the sum of cubes.

### Step-by-Step Solution:

1. **Identify the components**: Let \( a = \sin^2\left(\frac{\pi}{49}\right) \) and \( b = \cos^2\left(\frac{\pi}{49}\right) \). Then, we can rewrite the expression as:
\[ a^3 + b^3 - 1 + 3ab. \]

2. **Use the identity for the sum of cubes**: Recall that \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). We know that:
\[ a + b = \sin^2\left(\frac{\pi}{49}\right) + \cos^2\left(\frac{\pi}{49}\right) = 1. \]
Therefore, we can express \( a^3 + b^3 \) as:
\[ a^3 + b^3 = (1)(a^2 - ab + b^2) = a^2 - ab + b^2. \]

3. **Simplify \( a^2 + b^2 \)**: We can express \( a^2 + b^2 \) using the identity:
\[ a^2 + b^2 = (a + b)^2 - 2ab = 1 - 2ab. \]
Thus,
\[ a^3 + b^3 = (1 - 2ab) - ab = 1 - 3ab. \]

4. **Substitute back into the expression**: Now substitute \( a^3 + b^3 \) back into the original expression:
\[ (1 - 3ab) - 1 + 3ab. \]

5. **Combine like terms**: The expression simplifies to:
\[ 1 - 3ab - 1 + 3ab = 0. \]

6. **Final result**: Therefore, the value of the original expression is:
\[ 0. \]

### Conclusion:
The value of

\[ \sin^6\left(\frac{\pi}{49}\right) + \cos^6\left(\frac{\pi}{49}\right) - 1 + 3\sin^2\left(\frac{\pi}{49}\right)\cos^2\left(\frac{\pi}{49}\right) = 0. \]