The number of real roots of `(x+1/x)^3+x+1/x=0` is

Options

A. 0
B. 2
C. 4
D. 6

Answer

To find the number of real roots of the equation \((x + \frac{1}{x})^3 + (x + \frac{1}{x}) = 0\), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ (x + \frac{1}{x})^3 + (x + \frac{1}{x}) = 0 \]
Let \(y = x + \frac{1}{x}\). Then, we can rewrite the equation as:
\[ y^3 + y = 0 \]

### Step 2: Factor the Equation
Now, we can factor the equation:
\[ y(y^2 + 1) = 0 \]
This gives us two cases to consider:
1. \(y = 0\)
2. \(y^2 + 1 = 0\)

### Step 3: Solve the First Case
For the first case, \(y = 0\):
\[ x + \frac{1}{x} = 0 \]
Multiplying both sides by \(x\) (assuming \(x \neq 0\)):
\[ x^2 + 1 = 0 \]
This simplifies to:
\[ x^2 = -1 \]
Since the square of a real number cannot be negative, there are no real solutions from this case.

### Step 4: Solve the Second Case
For the second case, \(y^2 + 1 = 0\):
\[ y^2 = -1 \]
Again, since the square of a real number cannot be negative, there are no real solutions from this case either.

### Conclusion
Since both cases yield no real solutions, we conclude that the original equation has:
\[ \text{Number of real roots} = 0 \]

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