The number of roots of equation `((x-1)(x-3))/((x-2)(x-4))-e^(x)) ((x+1)(x+3)e^x)/((x+2)(x+4))-1) (x^(3)-cos x)=0`:

Answer

To determine the number of roots of the equation

\[ \frac{(x-1)(x-3)}{(x-2)(x-4)} - e^x \cdot \frac{(x+1)(x+3)e^x}{(x+2)(x+4)} - (x^3 - \cos x) = 0, \]

we will analyze the degree of the polynomial involved in the equation.

### Step-by-Step Solution:

1. **Identify the Components of the Equation**:
The equation consists of three main parts:
- The first term: \(\frac{(x-1)(x-3)}{(x-2)(x-4)}\)
- The second term: \(e^x \cdot \frac{(x+1)(x+3)e^x}{(x+2)(x+4)}\)
- The third term: \((x^3 - \cos x)\)

2. **Determine the Degree of Each Component**:
- For the first term, the numerator \((x-1)(x-3)\) has a degree of 2 (since it is a product of two linear factors), and the denominator \((x-2)(x-4)\) also has a degree of 2. Therefore, the degree of the first term is \(2 - 2 = 0\) (constant).
- For the second term, the numerator \(e^x \cdot (x+1)(x+3)e^x\) simplifies to \(e^{2x}(x^2 + 4x + 3)\). The polynomial part \((x+1)(x+3)\) has a degree of 2, and the exponential function does not contribute to the degree in the polynomial sense. The denominator \((x+2)(x+4)\) has a degree of 2. Thus, the degree of the second term is \(2 - 2 = 0\) (constant).
- For the third term, \(x^3 - \cos x\), the polynomial part \(x^3\) has a degree of 3. The \(-\cos x\) term is not a polynomial and does not affect the polynomial degree.

3. **Combine the Degrees**:
Since the first two terms are constants (degree 0), and the third term has a degree of 3, the overall degree of the equation is determined by the term with the highest degree, which is 3.

4. **Conclusion on the Number of Roots**:
A polynomial of degree \(n\) can have at most \(n\) roots. Therefore, since the highest degree we found is 3, the equation can have at most 3 roots.

### Final Answer:
The number of roots of the equation is **3**.