If ` x^(3) + y^(3) = 35 and x + y = 5` then the value of ` (1)/( x) + (1)/( y)`

Options

A. `(1)/(3)`
B. `(5)/(6)`
C. 6
D. `(2)/(3)`

Answer

To solve the problem, we need to find the value of \( \frac{1}{x} + \frac{1}{y} \) given the equations \( x^3 + y^3 = 35 \) and \( x + y = 5 \).

### Step-by-Step Solution:

1. **Use the identity for the sum of cubes**:
The identity for the sum of cubes states that:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
We know \( x + y = 5 \), so we can substitute this into the equation:
\[ x^3 + y^3 = 5(x^2 - xy + y^2) \]

2. **Substituting the known value**:
We know from the problem that \( x^3 + y^3 = 35 \). Therefore, we can set up the equation:
\[ 35 = 5(x^2 - xy + y^2) \]

3. **Simplifying the equation**:
Dividing both sides by 5 gives:
\[ x^2 - xy + y^2 = 7 \]

4. **Using the square of a sum**:
We can express \( x^2 + y^2 \) in terms of \( (x + y)^2 \) and \( xy \):
\[ x^2 + y^2 = (x + y)^2 - 2xy = 5^2 - 2xy = 25 - 2xy \]
Substituting this into our equation gives:
\[ 25 - 2xy - xy = 7 \]
Which simplifies to:
\[ 25 - 3xy = 7 \]

5. **Solving for \( xy \)**:
Rearranging the equation gives:
\[ 3xy = 25 - 7 \]
\[ 3xy = 18 \]
\[ xy = 6 \]

6. **Finding \( \frac{1}{x} + \frac{1}{y} \)**:
We know that:
\[ \frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy} \]
Substituting the known values:
\[ \frac{1}{x} + \frac{1}{y} = \frac{5}{6} \]

### Final Answer:
Thus, the value of \( \frac{1}{x} + \frac{1}{y} \) is \( \frac{5}{6} \).

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