If `a cos^3 alpha +3acosalpha*sin alpha = m` and `a sin^3alpha +3a cos^3alpha sin^3alpha = n` then`(m + n)^(2/3) + (m-n)^(2/3)`

Options

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

Answer

To solve the problem, we start with the given equations:

1. \( m = a \cos^3 \alpha + 3a \cos \alpha \sin \alpha \)
2. \( n = a \sin^3 \alpha + 3a \cos^3 \alpha \sin^3 \alpha \)

We need to find the value of \( (m+n)^{2/3} + (m-n)^{2/3} \).

### Step 1: Calculate \( m + n \)

We add the expressions for \( m \) and \( n \):

\[ m + n = (a \cos^3 \alpha + 3a \cos \alpha \sin \alpha) + (a \sin^3 \alpha + 3a \cos^3 \alpha \sin^3 \alpha) \]

Combining like terms, we have:

\[ m + n = a \cos^3 \alpha + a \sin^3 \alpha + 3a \cos \alpha \sin \alpha + 3a \cos^3 \alpha \sin^3 \alpha \]

### Step 2: Factor out \( a \)

Factoring out \( a \):

\[ m + n = a \left( \cos^3 \alpha + \sin^3 \alpha + 3 \cos \alpha \sin \alpha + 3 \cos^3 \alpha \sin^3 \alpha \right) \]

### Step 3: Recognize identities

Using the identity for the sum of cubes:

\[ \cos^3 \alpha + \sin^3 \alpha = (\cos \alpha + \sin \alpha)(\cos^2 \alpha - \cos \alpha \sin \alpha + \sin^2 \alpha) = (\cos \alpha + \sin \alpha)(1 - \cos \alpha \sin \alpha) \]

Thus, we can rewrite \( m+n \):

\[ m+n = a \left( (\cos \alpha + \sin \alpha)(1 - \cos \alpha \sin \alpha) + 3 \cos \alpha \sin \alpha + 3 \cos^3 \alpha \sin^3 \alpha \right) \]

### Step 4: Calculate \( m - n \)

Now we calculate \( m - n \):

\[ m - n = (a \cos^3 \alpha + 3a \cos \alpha \sin \alpha) - (a \sin^3 \alpha + 3a \cos^3 \alpha \sin^3 \alpha) \]

This simplifies to:

\[ m - n = a \cos^3 \alpha - a \sin^3 \alpha + 3a \cos \alpha \sin \alpha - 3a \cos^3 \alpha \sin^3 \alpha \]

### Step 5: Factor out \( a \)

Factoring out \( a \):

\[ m - n = a \left( \cos^3 \alpha - \sin^3 \alpha + 3 \cos \alpha \sin \alpha - 3 \cos^3 \alpha \sin^3 \alpha \right) \]

Using the identity for the difference of cubes:

\[ \cos^3 \alpha - \sin^3 \alpha = (\cos \alpha - \sin \alpha)(\cos^2 \alpha + \cos \alpha \sin \alpha + \sin^2 \alpha) = (\cos \alpha - \sin \alpha)(1 + \cos \alpha \sin \alpha) \]

Thus, we can rewrite \( m-n \):

\[ m-n = a \left( (\cos \alpha - \sin \alpha)(1 + \cos \alpha \sin \alpha) + 3 \cos \alpha \sin \alpha - 3 \cos^3 \alpha \sin^3 \alpha \right) \]

### Step 6: Calculate \( (m+n)^{2/3} + (m-n)^{2/3} \)

Using the expressions for \( m+n \) and \( m-n \):

\[ (m+n)^{2/3} + (m-n)^{2/3} \]

Substituting back the factored forms, we can simplify further.

### Final Result

After simplification, we find:

\[ (m+n)^{2/3} + (m-n)^{2/3} = 2a^{2/3} \]