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

Options

A. 2
B. 1
C. 0
D. None of these

Answer

To solve the expression \(\frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} + \frac{\cos^3 \theta - \sin^3 \theta}{\cos \theta - \sin \theta}\), we will use the identities for the sum and difference of cubes.

### Step-by-Step Solution:

1. **Identify the components of the expression**:
The expression consists of two parts:
\[ A = \frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} \]
\[ B = \frac{\cos^3 \theta - \sin^3 \theta}{\cos \theta - \sin \theta} \]

2. **Apply the sum of cubes formula**:
The sum of cubes formula states that:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Here, let \(a = \sin \theta\) and \(b = \cos \theta\):
\[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) \]
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we can simplify:
\[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \]

3. **Substitute back into \(A\)**:
Now substituting this back into \(A\):
\[ A = \frac{(\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta)}{\sin \theta + \cos \theta} \]
We can cancel \((\sin \theta + \cos \theta)\) (assuming \(\sin \theta + \cos \theta \neq 0\)):
\[ A = 1 - \sin \theta \cos \theta \]

4. **Apply the difference of cubes formula**:
The difference of cubes formula states that:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Here, we apply it to \(B\):
\[ \cos^3 \theta - \sin^3 \theta = (\cos \theta - \sin \theta)(\cos^2 \theta + \sin^2 \theta + \sin \theta \cos \theta) \]
Again, since \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ \cos^3 \theta - \sin^3 \theta = (\cos \theta - \sin \theta)(1 + \sin \theta \cos \theta) \]

5. **Substitute back into \(B\)**:
Now substituting this back into \(B\):
\[ B = \frac{(\cos \theta - \sin \theta)(1 + \sin \theta \cos \theta)}{\cos \theta - \sin \theta} \]
We can cancel \((\cos \theta - \sin \theta)\) (assuming \(\cos \theta - \sin \theta \neq 0\)):
\[ B = 1 + \sin \theta \cos \theta \]

6. **Combine \(A\) and \(B\)**:
Now we can combine \(A\) and \(B\):
\[ A + B = (1 - \sin \theta \cos \theta) + (1 + \sin \theta \cos \theta) \]
Simplifying this gives:
\[ A + B = 1 - \sin \theta \cos \theta + 1 + \sin \theta \cos \theta = 2 \]

### Final Answer:
The value of the expression is \(2\).