If `sin alpha -cos alpha = m`, then the value of `sin^6alpha+cos^6alpha` in terms of m is

Options

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

Answer

To find the value of \( \sin^6 \alpha + \cos^6 \alpha \) in terms of \( m \) where \( \sin \alpha - \cos \alpha = m \), we will follow these steps:

### Step 1: 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:
\[ \sin^6 \alpha + \cos^6 \alpha = (\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha + \cos^4 \alpha - \sin^2 \alpha \cos^2 \alpha) \]
Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we have:
\[ \sin^6 \alpha + \cos^6 \alpha = \sin^4 \alpha + \cos^4 \alpha - \sin^2 \alpha \cos^2 \alpha \]

### Step 2: Find \( \sin^4 \alpha + \cos^4 \alpha \)
We can use the identity for \( a^4 + b^4 \):
\[ \sin^4 \alpha + \cos^4 \alpha = (\sin^2 \alpha + \cos^2 \alpha)^2 - 2\sin^2 \alpha \cos^2 \alpha \]
Substituting \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ \sin^4 \alpha + \cos^4 \alpha = 1 - 2\sin^2 \alpha \cos^2 \alpha \]

### Step 3: Substitute back into the equation
Now substituting this back into our expression for \( \sin^6 \alpha + \cos^6 \alpha \):
\[ \sin^6 \alpha + \cos^6 \alpha = (1 - 2\sin^2 \alpha \cos^2 \alpha) - \sin^2 \alpha \cos^2 \alpha \]
This simplifies to:
\[ \sin^6 \alpha + \cos^6 \alpha = 1 - 3\sin^2 \alpha \cos^2 \alpha \]

### Step 4: Find \( \sin^2 \alpha \cos^2 \alpha \) in terms of \( m \)
From the given equation \( \sin \alpha - \cos \alpha = m \), we square both sides:
\[ \sin^2 \alpha - 2\sin \alpha \cos \alpha + \cos^2 \alpha = m^2 \]
Using \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ 1 - 2\sin \alpha \cos \alpha = m^2 \]
Thus, we can express \( \sin \alpha \cos \alpha \):
\[ \sin \alpha \cos \alpha = \frac{1 - m^2}{2} \]
Now squaring this gives:
\[ \sin^2 \alpha \cos^2 \alpha = \left(\frac{1 - m^2}{2}\right)^2 = \frac{(1 - m^2)^2}{4} \]

### Step 5: Substitute \( \sin^2 \alpha \cos^2 \alpha \) back into the equation
Substituting this back into our equation for \( \sin^6 \alpha + \cos^6 \alpha \):
\[ \sin^6 \alpha + \cos^6 \alpha = 1 - 3\left(\frac{(1 - m^2)^2}{4}\right) \]
This simplifies to:
\[ \sin^6 \alpha + \cos^6 \alpha = 1 - \frac{3(1 - m^2)^2}{4} \]
Expanding this gives:
\[ \sin^6 \alpha + \cos^6 \alpha = 1 - \frac{3(1 - 2m^2 + m^4)}{4} \]
\[ = 1 - \frac{3 - 6m^2 + 3m^4}{4} = \frac{4 - 3 + 6m^2 - 3m^4}{4} = \frac{1 + 6m^2 - 3m^4}{4} \]

### Final Result
Thus, we have:
\[ \sin^6 \alpha + \cos^6 \alpha = \frac{4 - 3(1 - m^2)^2}{4} \]