If `x + y = 4 and 1/x + 1/y = 16/15` find `x^(3) + y^(3)`.

Options

A. 19
B. 17
C. 15
D. 13

Answer

To solve the problem, we need to find \( x^3 + y^3 \) given the equations \( x + y = 4 \) and \( \frac{1}{x} + \frac{1}{y} = \frac{16}{15} \).

### Step-by-Step Solution:

1. **Use the given equations**:
We know that:
\[ x + y = 4 \]
and
\[ \frac{1}{x} + \frac{1}{y} = \frac{16}{15} \]

2. **Rewrite the second equation**:
The equation \( \frac{1}{x} + \frac{1}{y} \) can be rewritten using the formula:
\[ \frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy} \]
Thus, we can write:
\[ \frac{x + y}{xy} = \frac{16}{15} \]
Substituting \( x + y = 4 \):
\[ \frac{4}{xy} = \frac{16}{15} \]

3. **Solve for \( xy \)**:
Cross-multiplying gives:
\[ 4 \cdot 15 = 16 \cdot xy \]
Therefore:
\[ 60 = 16xy \]
Dividing both sides by 16:
\[ xy = \frac{60}{16} = \frac{15}{4} \]

4. **Use the formula for \( x^3 + y^3 \)**:
The formula for \( x^3 + y^3 \) is:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
We already have \( x + y = 4 \) and \( xy = \frac{15}{4} \).

5. **Calculate \( x^2 + y^2 \)**:
We can find \( x^2 + y^2 \) using the identity:
\[ x^2 + y^2 = (x + y)^2 - 2xy \]
Substituting the known values:
\[ x^2 + y^2 = 4^2 - 2 \cdot \frac{15}{4} = 16 - \frac{30}{4} = 16 - 7.5 = 8.5 = \frac{17}{2} \]

6. **Substitute into the formula**:
Now we can substitute \( x^2 + y^2 \) into the formula for \( x^3 + y^3 \):
\[ x^3 + y^3 = (x + y) \left( x^2 + y^2 - xy \right) \]
\[ = 4 \left( \frac{17}{2} - \frac{15}{4} \right) \]

7. **Simplify the expression**:
First, we need a common denominator to simplify:
\[ \frac{17}{2} = \frac{34}{4} \]
Therefore:
\[ x^2 + y^2 - xy = \frac{34}{4} - \frac{15}{4} = \frac{19}{4} \]
Now substituting back:
\[ x^3 + y^3 = 4 \cdot \frac{19}{4} = 19 \]

### Final Answer:
Thus, the value of \( x^3 + y^3 \) is \( \boxed{19} \).