If `sin A + cos A = p, sin^(3) A + cos^(3) A=q`, then

Options

A. `p^(3) - 3p + q =0`
B. `q^(3) - 3q + p =0`
C. `p^(3) - 3p + 2q =0`
D. ` q^(3) - 3q + 2p =0`

Answer

To solve the problem, we need to find a relationship between \( p \) and \( q \) given that \( \sin A + \cos A = p \) and \( \sin^3 A + \cos^3 A = q \).

### Step-by-step Solution:

1. **Start with the given equations**:
\[ \sin A + \cos A = p \]
\[ \sin^3 A + \cos^3 A = q \]

2. **Use the identity for the sum of cubes**:
The identity for the sum of cubes states that:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Here, let \( a = \sin A \) and \( b = \cos A \). Thus, we can write:
\[ \sin^3 A + \cos^3 A = (\sin A + \cos A)((\sin A)^2 - \sin A \cos A + (\cos A)^2) \]

3. **Substitute \( p \) into the equation**:
Since \( \sin A + \cos A = p \), we can substitute:
\[ q = p((\sin A)^2 - \sin A \cos A + (\cos A)^2) \]

4. **Simplify the expression**:
We know that:
\[ (\sin A)^2 + (\cos A)^2 = 1 \]
Therefore, we can rewrite the expression:
\[ q = p(1 - \sin A \cos A) \]

5. **Express \( \sin A \cos A \)**:
We can express \( \sin A \cos A \) in terms of \( p \):
\[ \sin A \cos A = \frac{1}{2} \sin(2A) \]
However, we can also use the identity:
\[ \sin A \cos A = \frac{p^2 - 1}{2} \]
This comes from the identity \( (\sin A + \cos A)^2 = \sin^2 A + \cos^2 A + 2\sin A \cos A \).

6. **Substituting back**:
Substitute \( \sin A \cos A \) back into the equation for \( q \):
\[ q = p\left(1 - \frac{p^2 - 1}{2}\right) \]
Simplifying this gives:
\[ q = p\left(1 - \frac{p^2}{2} + \frac{1}{2}\right) = p\left(\frac{3}{2} - \frac{p^2}{2}\right) \]

7. **Final expression**:
Thus, we have:
\[ q = \frac{p(3 - p^2)}{2} \]

### Conclusion:
The relationship between \( p \) and \( q \) is given by:
\[ q = \frac{p(3 - p^2)}{2} \]