The equation whose roots are multiplied by 3 of those of `2x^(3) - 3x^(2) + 4x - 5 = 0` is

Options

A. `2x^3 -9x^2 + 36 - 135=0`
B. `2x^3 -9x^2 -36 x + 135 =0`
C. `x^3 - 9x^2 + 36 x + 135 =0`
D. `2x^3 -9x^2 + 36 x + 135 =0`

Answer

To solve the problem, we need to find the equation whose roots are three times the roots of the given cubic equation \(2x^3 - 3x^2 + 4x - 5 = 0\).

### Step-by-Step Solution:

1. **Identify the coefficients of the given cubic equation**:
The given equation is:
\[ 2x^3 - 3x^2 + 4x - 5 = 0 \]
Here, \(a = 2\), \(b = -3\), \(c = 4\), and \(d = -5\).

2. **Calculate the sum of the roots**:
The sum of the roots \((\alpha + \beta + \gamma)\) can be calculated using the formula:
\[ \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-3}{2} = \frac{3}{2} \]

3. **Calculate the sum of the product of the roots taken two at a time**:
The sum of the product of the roots \((\alpha\beta + \beta\gamma + \gamma\alpha)\) is given by:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{4}{2} = 2 \]

4. **Calculate the product of the roots**:
The product of the roots \((\alpha \beta \gamma)\) can be calculated as:
\[ \alpha \beta \gamma = -\frac{d}{a} = -\frac{-5}{2} = \frac{5}{2} \]

5. **Formulate the new roots**:
We need to find the equation whose roots are \(3\alpha\), \(3\beta\), and \(3\gamma\).

6. **Calculate the new sum of the roots**:
The sum of the new roots is:
\[ 3\alpha + 3\beta + 3\gamma = 3(\alpha + \beta + \gamma) = 3 \times \frac{3}{2} = \frac{9}{2} \]

7. **Calculate the new sum of the product of the roots taken two at a time**:
The sum of the product of the new roots is:
\[ 3\alpha \cdot 3\beta + 3\beta \cdot 3\gamma + 3\gamma \cdot 3\alpha = 9(\alpha\beta + \beta\gamma + \gamma\alpha) = 9 \times 2 = 18 \]

8. **Calculate the new product of the roots**:
The product of the new roots is:
\[ 3\alpha \cdot 3\beta \cdot 3\gamma = 27(\alpha\beta\gamma) = 27 \times \frac{5}{2} = \frac{135}{2} \]

9. **Form the new cubic equation**:
The new cubic equation can be expressed as:
\[ x^3 - (\text{sum of roots})x^2 + (\text{sum of products of roots})x - (\text{product of roots}) = 0 \]
Substituting the values we calculated:
\[ x^3 - \frac{9}{2}x^2 + 18x - \frac{135}{2} = 0 \]

10. **Eliminate fractions by multiplying through by 2**:
\[ 2x^3 - 9x^2 + 36x - 135 = 0 \]

### Final Answer:
The equation whose roots are three times the roots of the given cubic equation is:
\[ 2x^3 - 9x^2 + 36x - 135 = 0 \]