If `alpha`, `beta`, `gamma` are the roots of the equation `9x^(3)-7x+6=0` then the equation `x^(3)+Ax^(2)+Bx+C=0` has roots `3alpha+2`, `3beta+2`, `3gamma+2`, where

Options

A. `A=6`
B. `B=-5`
C. `C=24`
D. `A+B+C=23`

Explanation

To solve the problem, we need to find the coefficients \( A \), \( B \), and \( C \) of the equation \( x^3 + Ax^2 + Bx + C = 0 \) whose roots are \( 3\alpha + 2 \), \( 3\beta + 2 \), and \( 3\gamma + 2 \), given that \( \alpha \), \( \beta \), and \( \gamma \) are the roots of the equation \( 9x^3 - 7x + 6 = 0 \).

### Step 1: Identify the roots of the original equation
The original equation is \( 9x^3 - 7x + 6 = 0 \). By Vieta's formulas, we know:
- \( \alpha + \beta + \gamma = 0 \) (coefficient of \( x^2 \) is 0)
- \( \alpha\beta + \beta\gamma + \gamma\alpha = -\frac{-7}{9} = \frac{7}{9} \)
- \( \alpha\beta\gamma = -\frac{6}{9} = -\frac{2}{3} \)

### Step 2: Find the new roots
The new roots are \( 3\alpha + 2 \), \( 3\beta + 2 \), and \( 3\gamma + 2 \).

### Step 3: Sum of the new roots
The sum of the new roots is:
\[ (3\alpha + 2) + (3\beta + 2) + (3\gamma + 2) = 3(\alpha + \beta + \gamma) + 6 = 3(0) + 6 = 6 \]

### Step 4: Product of the new roots taken two at a time
The product of the new roots taken two at a time is:
\[ (3\alpha + 2)(3\beta + 2) + (3\beta + 2)(3\gamma + 2) + (3\gamma + 2)(3\alpha + 2) \]
Expanding this:
\[ = 9(\alpha\beta + \beta\gamma + \gamma\alpha) + 6(\alpha + \beta + \gamma) + 4 \]
Substituting the values from Vieta's:
\[ = 9\left(\frac{7}{9}\right) + 6(0) + 4 = 7 + 0 + 4 = 11 \]

### Step 5: Product of the new roots
The product of the new roots is:
\[ (3\alpha + 2)(3\beta + 2)(3\gamma + 2) = 27\alpha\beta\gamma + 18(\alpha\beta + \beta\gamma + \gamma\alpha) + 12(\alpha + \beta + \gamma) + 8 \]
Substituting the values:
\[ = 27\left(-\frac{2}{3}\right) + 18\left(\frac{7}{9}\right) + 12(0) + 8 \]
Calculating this gives:
\[ = -18 + 14 + 0 + 8 = 4 \]

### Step 6: Form the new polynomial
Using the sums and products found:
- The sum of the roots \( = -A = 6 \) → \( A = -6 \)
- The sum of the products of the roots taken two at a time \( = B = 11 \)
- The product of the roots \( = -C = 4 \) → \( C = -4 \)

Thus, the coefficients are:
- \( A = -6 \)
- \( B = 11 \)
- \( C = -4 \)

### Final Answer
The values of \( A \), \( B \), and \( C \) are:
- \( A = -6 \)
- \( B = 11 \)
- \( C = -4 \)