If `alpha, beta gamma` are the real roots of the equation `x^(3)-3px^(2)+3qx-1=0`, then find the centroid of the triangle whose vertices are `(alpha, (1)/(alpha)), (beta, (1)/(beta))` and `(gamma, (1)/(gamma))`.

Answer

To find the centroid of the triangle whose vertices are \((\alpha, \frac{1}{\alpha})\), \((\beta, \frac{1}{\beta})\), and \((\gamma, \frac{1}{\gamma})\), where \(\alpha\), \(\beta\), and \(\gamma\) are the real roots of the equation \(x^3 - 3px^2 + 3qx - 1 = 0\), we can follow these steps:

### Step 1: Identify the formula for the centroid
The centroid \(G\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]

### Step 2: Substitute the vertices into the centroid formula
In our case, the vertices are:
- \((\alpha, \frac{1}{\alpha})\)
- \((\beta, \frac{1}{\beta})\)
- \((\gamma, \frac{1}{\gamma})\)

Thus, the coordinates of the centroid become:
\[ G = \left(\frac{\alpha + \beta + \gamma}{3}, \frac{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}}{3}\right) \]

### Step 3: Calculate \(\alpha + \beta + \gamma\)
Using Vieta's formulas for the polynomial \(x^3 - 3px^2 + 3qx - 1 = 0\):
- The sum of the roots \(\alpha + \beta + \gamma = 3p\).

### Step 4: Calculate \(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\)
Using the identity:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} \]
From Vieta's formulas:
- \(\beta\gamma + \gamma\alpha + \alpha\beta = 3q\)
- \(\alpha\beta\gamma = 1\)

Thus, we have:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{3q}{1} = 3q \]

### Step 5: Substitute back into the centroid formula
Now substituting back into the formula for the centroid:
\[ G = \left(\frac{3p}{3}, \frac{3q}{3}\right) = (p, q) \]

### Final Result
The centroid of the triangle is:
\[ \boxed{(p, q)} \]