If 1, -2, 3 are the roots of `ax^(3) + bx^(2) + cx + d = 0` then the roots of `ax^(3) + 3bx^(2) + 9cx + 27d = 0 ` are

Options

A. `2,4,6`
B. `3,-6,9`
C. `1/2 ,1,3/2`
D. `-1,0,1`

Answer

To solve the problem, we need to find the roots of the equation \( ax^3 + 3bx^2 + 9cx + 27d = 0 \) given that the roots of the equation \( ax^3 + bx^2 + cx + d = 0 \) are \( 1, -2, 3 \).

### Step 1: Understand the given roots
The roots of the polynomial \( ax^3 + bx^2 + cx + d = 0 \) are given as \( 1, -2, 3 \). This means we can express the polynomial as:
\[ f(x) = a(x - 1)(x + 2)(x - 3) \]

### Step 2: Substitute \( x \) with \( \frac{x}{3} \)
To find the roots of the new polynomial \( ax^3 + 3bx^2 + 9cx + 27d = 0 \), we will substitute \( x \) with \( \frac{x}{3} \) in the original polynomial \( f(x) \):
\[ f\left(\frac{x}{3}\right) = a\left(\frac{x}{3} - 1\right)\left(\frac{x}{3} + 2\right)\left(\frac{x}{3} - 3\right) \]

### Step 3: Simplify the expression
Now we simplify each term:
- \( \frac{x}{3} - 1 = \frac{x - 3}{3} \)
- \( \frac{x}{3} + 2 = \frac{x + 6}{3} \)
- \( \frac{x}{3} - 3 = \frac{x - 9}{3} \)

Thus, we can rewrite \( f\left(\frac{x}{3}\right) \) as:
\[ f\left(\frac{x}{3}\right) = a \cdot \frac{(x - 3)(x + 6)(x - 9)}{27} \]

### Step 4: Multiply by 27
To eliminate the fraction, we multiply the entire equation by 27:
\[ 27f\left(\frac{x}{3}\right) = a(x - 3)(x + 6)(x - 9) = 0 \]

### Step 5: Identify the new roots
The roots of the equation \( ax^3 + 3bx^2 + 9cx + 27d = 0 \) are the values of \( x \) that make \( (x - 3)(x + 6)(x - 9) = 0 \). Therefore, the roots are:
\[ x = 3, \quad x = -6, \quad x = 9 \]

### Final Answer
The roots of the equation \( ax^3 + 3bx^2 + 9cx + 27d = 0 \) are \( 3, -6, 9 \).

---