If `a+b+c=1`, `a^2+b^2+c^2=9` and `a^3+b^3+c^3=1`, then `1/a + 1/b + 1/c` is (i)`0` (ii)`-1` (iii)`1` (iv)`3`

Options

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

Answer

To solve the problem, we will use the given equations and some algebraic identities.

### Step 1: Write down the given equations
We have the following equations:
1. \( a + b + c = 1 \) (Equation 1)
2. \( a^2 + b^2 + c^2 = 9 \) (Equation 2)
3. \( a^3 + b^3 + c^3 = 1 \) (Equation 3)

### Step 2: Use the identity for the sum of squares
We know the identity:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) \]
Substituting the values from Equation 1 and Equation 2:
\[ 9 = (1)^2 - 2(ab + bc + ca) \]
This simplifies to:
\[ 9 = 1 - 2(ab + bc + ca) \]
Rearranging gives:
\[ 2(ab + bc + ca) = 1 - 9 = -8 \]
Thus,
\[ ab + bc + ca = -4 \quad \text{(Equation 4)} \]

### Step 3: Use the identity for the sum of cubes
We also have the identity:
\[ a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) + 3abc \]
Substituting the known values:
\[ 1 = (1)(9 - (-4)) + 3abc \]
This simplifies to:
\[ 1 = 1 \cdot (9 + 4) + 3abc \]
\[ 1 = 13 + 3abc \]
Rearranging gives:
\[ 3abc = 1 - 13 = -12 \]
Thus,
\[ abc = -4 \quad \text{(Equation 5)} \]

### Step 4: Find \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \)
Using the formula:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + ac + bc}{abc} \]
Substituting the values from Equation 4 and Equation 5:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{-4}{-4} = 1 \]

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

### Options
The correct option is (iii) \( 1 \).

---