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 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 equation \(x^3 + 3x + 5 = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The given cubic equation is:
\[ x^3 + 0x^2 + 3x + 5 = 0 \]
From this, we can identify:
- \(a = 1\)
- \(b = 0\)
- \(c = 3\)
- \(d = 5\)

### Step 2: Use Vieta's formulas
According to Vieta's formulas for a cubic equation \(ax^3 + bx^2 + cx + d = 0\):
- The sum of the roots \(x_1 + x_2 + x_3 = -\frac{b}{a} = -\frac{0}{1} = 0\)
- The sum of the product of the roots taken two at a time \(x_1x_2 + x_2x_3 + x_3x_1 = \frac{c}{a} = \frac{3}{1} = 3\)
- The product of the roots \(x_1x_2x_3 = -\frac{d}{a} = -\frac{5}{1} = -5\)

### Step 3: Rewrite the expression
We can rewrite the expression we want to evaluate:
\[ (x_1 + \frac{1}{x_1})(x_2 + \frac{1}{x_2})(x_3 + \frac{1}{x_3}) = (x_1 + \frac{1}{x_1})(x_2 + \frac{1}{x_2})(x_3 + \frac{1}{x_3}) \]
This can be simplified to:
\[ = (x_1 + \frac{1}{x_1})(x_2 + \frac{1}{x_2})(x_3 + \frac{1}{x_3}) = (x_1 + \frac{1}{x_1})(x_2 + \frac{1}{x_2})(x_3 + \frac{1}{x_3}) \]

### Step 4: Find the individual terms
Each term can be expressed as:
\[ x_i + \frac{1}{x_i} = \frac{x_i^2 + 1}{x_i} \]
Thus, the product becomes:
\[ P = \frac{(x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)}{x_1x_2x_3} \]

### Step 5: Calculate \(x_1^2 + x_2^2 + x_3^2\)
Using 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 we found:
\[ x_1^2 + x_2^2 + x_3^2 = 0^2 - 2 \cdot 3 = -6 \]

### Step 6: Calculate \(x_1^2 + 1\), \(x_2^2 + 1\), \(x_3^2 + 1\)
Now we can find:
\[ (x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1) = (x_1^2 + x_2^2 + x_3^2) + 3 + (x_1^2x_2^2 + x_2^2x_3^2 + x_3^2x_1^2) \]
We already have \(x_1^2 + x_2^2 + x_3^2 = -6\).

### Step 7: Calculate \(x_1^2x_2^2 + x_2^2x_3^2 + x_3^2x_1^2\)
Using:
\[ x_1^2x_2^2 + x_2^2x_3^2 + x_3^2x_1^2 = (x_1x_2 + x_2x_3 + x_3x_1)^2 - 2x_1x_2x_3(x_1 + x_2 + x_3) \]
Substituting the values:
\[ = 3^2 - 2(-5)(0) = 9 \]

### Step 8: Substitute back into \(P\)
Now we can substitute back into our expression for \(P\):
\[ P = \frac{(-6 + 3 + 9)}{-5} = \frac{6}{-5} = -\frac{6}{5} \]

### 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:
\[ \boxed{-\frac{6}{5}} \]