The value of `(cos^(3)theta+sin^(3)theta)/(costheta+sintheta)+(cos^(3)theta-sin^(3)theta)/(costheta-sintheta)` is equal to

Options

A. `-1`
B. 1
C. 2
D. 0

Answer

To solve the expression

\[ \frac{\cos^3 \theta + \sin^3 \theta}{\cos \theta + \sin \theta} + \frac{\cos^3 \theta - \sin^3 \theta}{\cos \theta - \sin \theta}, \]

we can use the identities for the sum and difference of cubes.

### Step 1: Apply the sum of cubes formula
Recall that the sum of cubes can be factored as:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2). \]

Let \( A = \cos \theta \) and \( B = \sin \theta \). Then we can write:

\[ \cos^3 \theta + \sin^3 \theta = (A + B)(A^2 - AB + B^2). \]

### Step 2: Substitute into the expression
Now substituting this into the first part of the expression:

\[ \frac{(A + B)(A^2 - AB + B^2)}{A + B}. \]

Since \( A + B \) is not zero (assuming \( \theta \) is not \( \frac{\pi}{4} + n\frac{\pi}{2} \)), we can simplify:

\[ A^2 - AB + B^2. \]

### Step 3: Apply the difference of cubes formula
For the second part, we use the difference of cubes formula:

\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2). \]

Thus,

\[ \cos^3 \theta - \sin^3 \theta = (A - B)(A^2 + AB + B^2). \]

Substituting this into the second part of the expression gives:

\[ \frac{(A - B)(A^2 + AB + B^2)}{A - B}. \]

Again, assuming \( A - B \) is not zero (which is true unless \( \theta \) is \( \frac{\pi}{4} + n\frac{\pi}{2} \)), we can simplify:

\[ A^2 + AB + B^2. \]

### Step 4: Combine the results
Now our expression simplifies to:

\[ (A^2 - AB + B^2) + (A^2 + AB + B^2). \]

Combining these gives:

\[ 2A^2 + 2B^2 = 2(\cos^2 \theta + \sin^2 \theta). \]

### Step 5: Use the Pythagorean identity
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ 2(\cos^2 \theta + \sin^2 \theta) = 2 \cdot 1 = 2. \]

### Final Answer
Thus, the value of the original expression is:

\[ \boxed{2}. \]