let f(x) be a polynomial satisfying f(x) : f(1/x) = f(x) + f(1/x) for all `X in R` :- {O} and f(5) =126, then find f(3).

Answer

To solve the problem, we need to find the polynomial \( f(x) \) that satisfies the condition \( f(x) \cdot f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right) \) for all \( x \in \mathbb{R} \setminus \{0\} \), and also satisfies \( f(5) = 126 \). We will then use this polynomial to find \( f(3) \).

### Step-by-Step Solution:

1. **Assume the Form of the Polynomial:**
We assume that \( f(x) \) can be expressed in the form \( f(x) = x^n + c \), where \( n \) is a non-negative integer and \( c \) is a constant.

2. **Substituting into the Given Condition:**
Substitute \( f(x) \) and \( f\left(\frac{1}{x}\right) \) into the condition:
\[ f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^n + c = \frac{1}{x^n} + c \]
Now, substituting into the equation:
\[ f(x) \cdot f\left(\frac{1}{x}\right) = \left(x^n + c\right)\left(\frac{1}{x^n} + c\right) \]
Expanding this gives:
\[ = x^n \cdot \frac{1}{x^n} + cx^n + c\frac{1}{x^n} + c^2 = 1 + cx^n + \frac{c}{x^n} + c^2 \]

3. **Finding \( f(x) + f\left(\frac{1}{x}\right) \):**
Now, calculate \( f(x) + f\left(\frac{1}{x}\right) \):
\[ f(x) + f\left(\frac{1}{x}\right) = \left(x^n + c\right) + \left(\frac{1}{x^n} + c\right) = x^n + \frac{1}{x^n} + 2c \]

4. **Setting the Two Expressions Equal:**
Set the two expressions equal to each other:
\[ 1 + cx^n + \frac{c}{x^n} + c^2 = x^n + \frac{1}{x^n} + 2c \]

5. **Equating Coefficients:**
For the equation to hold for all \( x \), the coefficients of \( x^n \) and \( \frac{1}{x^n} \) must match. This gives us:
- From \( cx^n = x^n \), we find \( c = 1 \).
- From \( \frac{c}{x^n} = \frac{1}{x^n} \), we confirm \( c = 1 \).

6. **Finding the Polynomial:**
Thus, we have:
\[ f(x) = x^n + 1 \]
Since \( c = 1 \) and we need to determine \( n \).

7. **Using the Given Condition \( f(5) = 126 \):**
Substitute \( x = 5 \):
\[ f(5) = 5^n + 1 = 126 \]
This simplifies to:
\[ 5^n = 125 \implies 5^n = 5^3 \implies n = 3 \]

8. **Final Form of the Polynomial:**
Therefore, the polynomial is:
\[ f(x) = x^3 + 1 \]

9. **Finding \( f(3) \):**
Now, we substitute \( x = 3 \):
\[ f(3) = 3^3 + 1 = 27 + 1 = 28 \]

### Final Answer:
Thus, \( f(3) = 28 \).