If `x+ y = 3`, then what is the value of `x^(3) +y^(3) + 9xy` ?

Options

A. 15
B. 81
C. 27
D. 9

Answer

To find the value of \( x^3 + y^3 + 9xy \) given that \( x + y = 3 \), we can use the identity for the sum of cubes and some algebraic manipulation.

### Step-by-step Solution:

1. **Use the identity for the sum of cubes**:
The identity states that:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
We can also express \( x^2 - xy + y^2 \) in terms of \( x + y \) and \( xy \).

2. **Express \( x^2 + y^2 \)**:
We know that:
\[ x^2 + y^2 = (x + y)^2 - 2xy \]
Substituting \( x + y = 3 \):
\[ x^2 + y^2 = 3^2 - 2xy = 9 - 2xy \]

3. **Substituting back into the identity**:
Now substituting \( x^2 + y^2 \) into the identity for \( x^3 + y^3 \):
\[ x^3 + y^3 = (x + y)((x^2 + y^2) - xy) = (x + y)((9 - 2xy) - xy) = (x + y)(9 - 3xy) \]
Since \( x + y = 3 \):
\[ x^3 + y^3 = 3(9 - 3xy) = 27 - 9xy \]

4. **Now, find \( x^3 + y^3 + 9xy \)**:
We need to find:
\[ x^3 + y^3 + 9xy = (27 - 9xy) + 9xy \]
Simplifying this gives:
\[ x^3 + y^3 + 9xy = 27 \]

Thus, the value of \( x^3 + y^3 + 9xy \) is \( \boxed{27} \).