If `alpha,beta,gamma` are roots of equation `x^3-6x^2+10x-3=0` then find equation with roots `2alpha+1, 2beta+1, 2gamma+1`

Explanation

To find the equation with roots \(2\alpha + 1\), \(2\beta + 1\), and \(2\gamma + 1\) where \(\alpha\), \(\beta\), and \(\gamma\) are the roots of the polynomial \(x^3 - 6x^2 + 10x - 3 = 0\), we can follow these steps:

### Step 1: Identify the coefficients of the given polynomial
The polynomial is given as:
\[ x^3 - 6x^2 + 10x - 3 = 0 \]
From this, we can identify:
- \(a = 1\)
- \(b = -6\)
- \(c = 10\)
- \(d = -3\)

### Step 2: Calculate the sums and products of the roots
Using Vieta's formulas:
- The sum of the roots \(\alpha + \beta + \gamma\) is given by:
\[ \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-6}{1} = 6 \]
- The sum of the products of the roots taken two at a time \(\alpha\beta + \beta\gamma + \gamma\alpha\) is:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{10}{1} = 10 \]
- The product of the roots \(\alpha\beta\gamma\) is:
\[ \alpha\beta\gamma = -\frac{d}{a} = -\frac{-3}{1} = 3 \]

### Step 3: Find the new roots
We need to find the new roots \(2\alpha + 1\), \(2\beta + 1\), and \(2\gamma + 1\).

#### Sum of the new roots
\[ (2\alpha + 1) + (2\beta + 1) + (2\gamma + 1) = 2(\alpha + \beta + \gamma) + 3 = 2 \cdot 6 + 3 = 12 + 3 = 15 \]

#### Sum of the products of the new roots taken two at a time
\[ (2\alpha + 1)(2\beta + 1) + (2\beta + 1)(2\gamma + 1) + (2\gamma + 1)(2\alpha + 1) \]
Expanding this:
\[ = 4(\alpha\beta + \beta\gamma + \gamma\alpha) + 2(\alpha + \beta + \gamma) + 3 \]
Substituting the values:
\[ = 4 \cdot 10 + 2 \cdot 6 + 3 = 40 + 12 + 3 = 55 \]

#### Product of the new roots
\[ (2\alpha + 1)(2\beta + 1)(2\gamma + 1) \]
Expanding this:
\[ = 8\alpha\beta\gamma + 4(\alpha\beta + \beta\gamma + \gamma\alpha) + 2(\alpha + \beta + \gamma) + 1 \]
Substituting the values:
\[ = 8 \cdot 3 + 4 \cdot 10 + 2 \cdot 6 + 1 = 24 + 40 + 12 + 1 = 77 \]

### Step 4: Form the new polynomial
Using the new sums and products, we can form the polynomial:
\[ x^3 - (\text{sum of roots})x^2 + (\text{sum of products of roots})x - (\text{product of roots}) = 0 \]
Substituting the values:
\[ x^3 - 15x^2 + 55x - 77 = 0 \]

### Final Answer
The equation with roots \(2\alpha + 1\), \(2\beta + 1\), and \(2\gamma + 1\) is:
\[ \boxed{x^3 - 15x^2 + 55x - 77 = 0} \]