What is the value of `(sin^(3) x + cos^(3) x)/( sin x + cos x) + sin x cos x ` ?

Options

A. 0
B. sin x
C. cos x
D. 1

Answer

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

### Step 1: Recognize the identity for the sum of cubes
The expression \(\sin^3 x + \cos^3 x\) can be rewritten using the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Here, let \(a = \sin x\) and \(b = \cos x\).

### Step 2: Apply the identity
Using the identity, we can rewrite \(\sin^3 x + \cos^3 x\):
\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x) \]

### Step 3: Substitute back into the expression
Now, substituting this back into our original expression:
\[ \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} + \sin x \cos x = \frac{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{\sin x + \cos x} + \sin x \cos x \]

### Step 4: Simplify the fraction
The \((\sin x + \cos x)\) terms in the numerator and denominator cancel out (as long as \(\sin x + \cos x \neq 0\)):
\[ \sin^2 x - \sin x \cos x + \cos^2 x + \sin x \cos x \]

### Step 5: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \sin^2 x + \cos^2 x = 1 \]
So we can substitute this into our expression:
\[ 1 - \sin x \cos x + \sin x \cos x \]

### Step 6: Combine like terms
Now, the \(-\sin x \cos x\) and \(+\sin x \cos x\) cancel each other out:
\[ 1 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{1} \]