If `alpha,beta, gamma` are the real roots of the equation `x^(3)-3ax^(2)+3bx-1=0`, then the centroid of the triangle whose vertices are the points `(alpha, 1/(alpha)), (beta, 1/(beta))`and `(gamma,1/(gamma))` is the point

Options

A. `(a,b)`
B. `(a,-b)`
C. `(-a,b)`
D. `(-a,-b)`

Answer

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

### Step 1: Identify the coordinates of the vertices
The vertices of the triangle are given as:
- \(A(\alpha, \frac{1}{\alpha})\)
- \(B(\beta, \frac{1}{\beta})\)
- \(C(\gamma, \frac{1}{\gamma})\)

### Step 2: Calculate the coordinates of the centroid
The formula for the centroid \(G\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
In our case, we have:
\[ G\left(\frac{\alpha + \beta + \gamma}{3}, \frac{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}}{3}\right) \]

### Step 3: Use Vieta's formulas
From Vieta's formulas for the polynomial \(x^3 - 3ax^2 + 3bx - 1 = 0\):
- The sum of the roots \(\alpha + \beta + \gamma = 3a\)
- The sum of the product of the roots taken two at a time \(\alpha\beta + \beta\gamma + \gamma\alpha = 3b\)
- The product of the roots \(\alpha\beta\gamma = 1\)

### Step 4: Substitute the values into the centroid formula
Now substituting the values into the centroid formula:
1. For the x-coordinate:
\[ \frac{\alpha + \beta + \gamma}{3} = \frac{3a}{3} = a \]

2. For the y-coordinate:
\[ \frac{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}}{3} = \frac{\frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma}}{3} = \frac{3b}{3 \cdot 1} = b \]

### Step 5: Write the final coordinates of the centroid
Thus, the coordinates of the centroid \(G\) are:
\[ G(a, b) \]

### Final Answer:
The centroid of the triangle is the point \((a, b)\).

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