Simplify : `( sin^(3) theta + cos ^(3) theta )/( sin theta+ cos theta ) + sin theta cos theta `

Options

A. 0
B. 1
C. `sin ^(2) theta - cos ^(2) theta `
D. `sin theta - cos theta `

Answer

To simplify the expression \(( \sin^3 \theta + \cos^3 \theta ) / ( \sin \theta + \cos \theta ) + \sin \theta \cos \theta\), we can follow these steps:

### Step 1: Apply the sum of cubes formula
The sum of cubes can be factored using the formula:
\[ A^3 + B^3 = (A + B)(A^2 - AB + B^2) \]
In our case, let \(A = \sin \theta\) and \(B = \cos \theta\). Therefore:
\[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) \]

### Step 2: Substitute the factorization into the expression
Now we can substitute this back into our original expression:
\[ \frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} + \sin \theta \cos \theta = \frac{(\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta + \cos \theta} + \sin \theta \cos \theta \]

### Step 3: Cancel out the common terms
Since \(\sin \theta + \cos \theta\) is in both the numerator and denominator, we can cancel it out (assuming \(\sin \theta + \cos \theta \neq 0\)):
\[ \sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta + \sin \theta \cos \theta \]

### Step 4: Simplify the expression
Now, we know that \(\sin^2 \theta + \cos^2 \theta = 1\). Therefore:
\[ 1 - \sin \theta \cos \theta + \sin \theta \cos \theta = 1 \]

### Final Result
Thus, the simplified expression is:
\[ \boxed{1} \]