If `a+b+c=6and(1)/(a)+(1)/(b)+(1)/(c)=(3)/(2)`, then find `(a)/(b)+(a)/(c)+(b)/(a)+(b)/(c)+(c)/(a)+(c)/(b)`.

Options

A. 6
B. 4
C. 9
D. 12

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations:**
\[ a + b + c = 6 \]
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{3}{2} \]

2. **Rearranging the Second Equation:**
We can rewrite the second equation using a common denominator:
\[ \frac{bc + ac + ab}{abc} = \frac{3}{2} \]
This implies:
\[ 2(bc + ac + ab) = 3abc \]

3. **Substituting \(a + b + c\):**
We know \(a + b + c = 6\). We can express \(bc + ac + ab\) in terms of \(a + b + c\) and \(abc\):
\[ bc + ac + ab = \frac{3}{2} \cdot \frac{abc}{2} \]
Rearranging gives:
\[ bc + ac + ab = \frac{3abc}{4} \]

4. **Finding the Expression:**
We need to find:
\[ \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} \]
This can be rewritten as:
\[ \left(\frac{a}{b} + \frac{b}{a}\right) + \left(\frac{a}{c} + \frac{c}{a}\right) + \left(\frac{b}{c} + \frac{c}{b}\right) \]
Each pair can be expressed as:
\[ \frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab} \]
Thus, we have:
\[ \frac{a^2 + b^2}{ab} + \frac{a^2 + c^2}{ac} + \frac{b^2 + c^2}{bc} \]

5. **Using the Identity:**
We can use the identity:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Substituting \(a + b + c = 6\):
\[ 36 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Rearranging gives:
\[ a^2 + b^2 + c^2 = 36 - 2(ab + ac + bc) \]

6. **Substituting Back:**
We can substitute \(ab + ac + bc\) from earlier:
\[ ab + ac + bc = \frac{3abc}{4} \]
Thus:
\[ a^2 + b^2 + c^2 = 36 - 2 \cdot \frac{3abc}{4} \]

7. **Final Calculation:**
Now we can substitute back into our expression:
\[ \frac{a^2 + b^2 + c^2}{ab} + \frac{a^2 + b^2 + c^2}{ac} + \frac{a^2 + b^2 + c^2}{bc} \]
After simplification, we find:
\[ \frac{(a^2 + b^2 + c^2) + 3(ab + ac + bc)}{abc} = 6 \]

Thus, the final answer is:
\[ \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} = 6 \]