Let `f: R - {0} to R` be defined by `f(x) = x+ 1/x`, then
`7 + f((x))^(4) -f(x^(4)) - 4(f(x))^(2)` is equal to……….

Answer

To solve the problem, we need to evaluate the expression \( 7 + f(x)^4 - f(x^4) - 4f(x)^2 \) where \( f(x) = x + \frac{1}{x} \).

### Step-by-Step Solution:

1. **Calculate \( f(x) \)**:
\[ f(x) = x + \frac{1}{x} \]

2. **Calculate \( f(x^4) \)**:
\[ f(x^4) = x^4 + \frac{1}{x^4} \]

3. **Calculate \( f(x)^2 \)**:
\[ f(x)^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \]

4. **Calculate \( f(x)^4 \)**:
\[ f(x)^4 = \left(f(x)^2\right)^2 = \left(x^2 + 2 + \frac{1}{x^2}\right)^2 \]
Expanding this:
\[ = x^4 + 4 + \frac{1}{x^4} + 4\left(x^2 + \frac{1}{x^2}\right) \]
\[ = x^4 + 4 + \frac{1}{x^4} + 4\left(x^2 + \frac{1}{x^2}\right) \]

5. **Substitute into the expression**:
We need to substitute \( f(x)^4 \) and \( f(x^4) \) into the expression:
\[ 7 + f(x)^4 - f(x^4) - 4f(x)^2 \]
This becomes:
\[ 7 + \left(x^4 + 4 + \frac{1}{x^4} + 4\left(x^2 + \frac{1}{x^2}\right)\right) - \left(x^4 + \frac{1}{x^4}\right) - 4\left(x^2 + 2 + \frac{1}{x^2}\right) \]

6. **Simplify the expression**:
\[ = 7 + x^4 + 4 + \frac{1}{x^4} + 4\left(x^2 + \frac{1}{x^2}\right) - x^4 - \frac{1}{x^4} - 4x^2 - 8 - 4\frac{1}{x^2} \]
The \( x^4 \) and \( \frac{1}{x^4} \) terms cancel out:
\[ = 7 + 4 - 8 + 4\left(x^2 + \frac{1}{x^2}\right) - 4x^2 - 4\frac{1}{x^2} \]
\[ = 3 + 4 - 8 = -1 \]

### Final Result:
Thus, the expression evaluates to:
\[ \boxed{-1} \]