If `sin alpha+ cos alpha=n`, then ` sin^(6)alpha+ cos^(6)alpha` in terms of n is ________

Options

A. `4+3(n-1)^(2)`
B. `(4+3(n^(2)-1))/(4)`
C. `4-3(n^(2)-1)^(2))/(4)`
D. None of these

Explanation

To solve the problem, we need to find \( \sin^6 \alpha + \cos^6 \alpha \) in terms of \( n \), given that \( \sin \alpha + \cos \alpha = n \).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \sin \alpha + \cos \alpha = n \]

2. **Use the identity for \( \sin^6 \alpha + \cos^6 \alpha \):**
We can express \( \sin^6 \alpha + \cos^6 \alpha \) using the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Here, let \( a = \sin^2 \alpha \) and \( b = \cos^2 \alpha \). Thus,
\[ \sin^6 \alpha + \cos^6 \alpha = (\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha) \]

3. **Use the Pythagorean identity:**
Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we have:
\[ \sin^6 \alpha + \cos^6 \alpha = 1 \cdot (\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha) \]
Therefore,
\[ \sin^6 \alpha + \cos^6 \alpha = \sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha \]

4. **Express \( \sin^4 \alpha \) and \( \cos^4 \alpha \):**
We can use the square of \( \sin^2 \alpha + \cos^2 \alpha \):
\[ \sin^4 \alpha + \cos^4 \alpha = (\sin^2 \alpha + \cos^2 \alpha)^2 - 2\sin^2 \alpha \cos^2 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha \]

5. **Substitute back into the equation:**
Now substituting back, we get:
\[ \sin^6 \alpha + \cos^6 \alpha = (1 - 2\sin^2 \alpha \cos^2 \alpha) - \sin^2 \alpha \cos^2 \alpha \]
This simplifies to:
\[ \sin^6 \alpha + \cos^6 \alpha = 1 - 3\sin^2 \alpha \cos^2 \alpha \]

6. **Find \( \sin^2 \alpha \cos^2 \alpha \):**
We know:
\[ \sin \alpha + \cos \alpha = n \implies (\sin \alpha + \cos \alpha)^2 = n^2 \]
Expanding this gives:
\[ \sin^2 \alpha + \cos^2 \alpha + 2\sin \alpha \cos \alpha = n^2 \]
Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ 1 + 2\sin \alpha \cos \alpha = n^2 \implies 2\sin \alpha \cos \alpha = n^2 - 1 \]
Thus,
\[ \sin \alpha \cos \alpha = \frac{n^2 - 1}{2} \]

7. **Substitute \( \sin^2 \alpha \cos^2 \alpha \):**
Therefore,
\[ \sin^2 \alpha \cos^2 \alpha = \left(\frac{n^2 - 1}{2}\right)^2 = \frac{(n^2 - 1)^2}{4} \]

8. **Final substitution:**
Substitute this back into our expression for \( \sin^6 \alpha + \cos^6 \alpha \):
\[ \sin^6 \alpha + \cos^6 \alpha = 1 - 3 \cdot \frac{(n^2 - 1)^2}{4} \]
Simplifying this gives:
\[ \sin^6 \alpha + \cos^6 \alpha = 1 - \frac{3(n^2 - 1)^2}{4} \]

9. **Final answer:**
Thus, we can express \( \sin^6 \alpha + \cos^6 \alpha \) in terms of \( n \):
\[ \sin^6 \alpha + \cos^6 \alpha = \frac{4 - 3(n^2 - 1)^2}{4} \]