If ` a^(3) + b^(3) = 9 and a + b = 3 ` then the value of ` (1)/( a) + (1)/( b)` is

Options

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

Answer

To solve the problem, we need to find the value of \( \frac{1}{a} + \frac{1}{b} \) given that \( a^3 + b^3 = 9 \) and \( a + b = 3 \).

### Step 1: Use the identity for the sum of cubes
We know that:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
We can substitute \( a + b = 3 \) into the equation:
\[ a^3 + b^3 = 3(a^2 - ab + b^2) \]

### Step 2: Substitute the value of \( a^3 + b^3 \)
Given \( a^3 + b^3 = 9 \), we can set up the equation:
\[ 9 = 3(a^2 - ab + b^2) \]
Dividing both sides by 3 gives:
\[ a^2 - ab + b^2 = 3 \]

### Step 3: Express \( a^2 + b^2 \) in terms of \( ab \)
We can use the identity:
\[ a^2 + b^2 = (a + b)^2 - 2ab \]
Substituting \( a + b = 3 \):
\[ a^2 + b^2 = 3^2 - 2ab = 9 - 2ab \]
Now we can substitute \( a^2 + b^2 \) into the equation from Step 2:
\[ (9 - 2ab) - ab = 3 \]
This simplifies to:
\[ 9 - 3ab = 3 \]

### Step 4: Solve for \( ab \)
Rearranging gives:
\[ 9 - 3 = 3ab \implies 6 = 3ab \implies ab = 2 \]

### Step 5: Find \( \frac{1}{a} + \frac{1}{b} \)
We know that:
\[ \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} \]
Substituting the values we have:
\[ \frac{1}{a} + \frac{1}{b} = \frac{3}{2} \]

### Final Answer
Thus, the value of \( \frac{1}{a} + \frac{1}{b} \) is \( \frac{3}{2} \).

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