If `(x+1/(x))^(2)=3`, show that `x^(3)+1/(x^(3))=0`

Answer

To solve the problem, we start with the equation given:

\[ \left(x + \frac{1}{x}\right)^2 = 3 \]

### Step 1: Simplify the equation
First, we take the square root of both sides:

\[ x + \frac{1}{x} = \sqrt{3} \quad \text{or} \quad x + \frac{1}{x} = -\sqrt{3} \]

### Step 2: Cube the expression
Next, we will cube the expression \(x + \frac{1}{x}\):

\[ \left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) \]

### Step 3: Substitute the value
Now, we substitute \(x + \frac{1}{x} = \sqrt{3}\) into the cubed expression:

\[ \left(\sqrt{3}\right)^3 = x^3 + \frac{1}{x^3} + 3\sqrt{3} \]

Calculating \(\left(\sqrt{3}\right)^3\):

\[ 3\sqrt{3} = x^3 + \frac{1}{x^3} + 3\sqrt{3} \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation to isolate \(x^3 + \frac{1}{x^3}\):

\[ x^3 + \frac{1}{x^3} = 3\sqrt{3} - 3\sqrt{3} \]

This simplifies to:

\[ x^3 + \frac{1}{x^3} = 0 \]

### Conclusion
Thus, we have shown that:

\[ x^3 + \frac{1}{x^3} = 0 \]