Let `x_(1),x_(2), x_(3)` be roots of equation `x^3 + 3x + 5 = 0`. What is the value of the expression `(x_(1) + 1/x_(1))(x_(2)+1/x_(2)) (x_(3)+1/x_(3))` ?

Answer

To solve the problem, we need to find the value of the expression \((x_1 + \frac{1}{x_1})(x_2 + \frac{1}{x_2})(x_3 + \frac{1}{x_3})\) where \(x_1, x_2, x_3\) are the roots of the polynomial equation \(x^3 + 3x + 5 = 0\).

### Step 1: Identify the roots' relationships
From Vieta's formulas for the polynomial \(x^3 + 0x^2 + 3x + 5 = 0\):
- The sum of the roots \(x_1 + x_2 + x_3 = 0\) (coefficient of \(x^2\) is 0).
- The sum of the products of the roots taken two at a time \(x_1x_2 + x_2x_3 + x_3x_1 = 3\) (coefficient of \(x\)).
- The product of the roots \(x_1x_2x_3 = -5\) (negative of the constant term).

### Step 2: Rewrite the expression
The expression can be rewritten using the identity:
\[ x + \frac{1}{x} = \frac{x^2 + 1}{x} \]
Thus, we have:
\[ (x_1 + \frac{1}{x_1})(x_2 + \frac{1}{x_2})(x_3 + \frac{1}{x_3}) = \frac{(x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)}{x_1 x_2 x_3} \]

### Step 3: Calculate \(x_1^2 + 1\), \(x_2^2 + 1\), \(x_3^2 + 1\)
We need to find \(x_1^2 + x_2^2 + x_3^2\). We can use the identity:
\[ x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2 - 2(x_1x_2 + x_2x_3 + x_3x_1) \]
Substituting the values:
\[ x_1^2 + x_2^2 + x_3^2 = 0^2 - 2 \cdot 3 = -6 \]

### Step 4: Calculate \(x_1^2 + 1\), \(x_2^2 + 1\), \(x_3^2 + 1\)
Now, we can find:
\[ x_1^2 + 1 = x_1^2 + x_2^2 + x_3^2 + 3 = -6 + 3 = -3 \]
Thus:
\[ (x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1) = (-3)(-3)(-3) = -27 \]

### Step 5: Substitute into the expression
Now substituting back into our expression:
\[ \frac{(x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)}{x_1 x_2 x_3} = \frac{-27}{-5} = \frac{27}{5} \]

### Final Answer
Thus, the value of the expression \((x_1 + \frac{1}{x_1})(x_2 + \frac{1}{x_2})(x_3 + \frac{1}{x_3})\) is:
\[ \boxed{\frac{27}{5}} \]