If the roots of the equation `a x^2-b x+c=0a r ealpha,beta,` then the roots of the equation `b^2c x^2-a b^(2x)+a^3=0` are `1/(alpha^3+alphabeta),1/(beta^3+alphabeta)` b. `1/(alpha^2+alphabeta),1/(beta^2+alphabeta)` c. `1/(alpha^4+alphabeta),1/(beta^4+alphabeta)` d. none of these

Options

A. `(1)/(alpha^(3) + alphabeta), (1)/(beta^(3) + alphabeta)`
B. `(1)/(alpha^(2) + alphabeta), (1)/(beta^(2) + alphabeta)`
C. `(1)/(alpha^(4) + alphabeta), (1)/(beta^(4) + alphabeta)`
D. none of these

Explanation

To solve the problem, we need to find the roots of the equation \( b^2c x^2 - ab^2x + a^3 = 0 \) given that the roots of the equation \( ax^2 - bx + c = 0 \) are \( \alpha \) and \( \beta \).

### Step 1: Understand the given equations
The first equation is:
\[ ax^2 - bx + c = 0 \]
with roots \( \alpha \) and \( \beta \).

From Vieta's formulas, we know:
- \( \alpha + \beta = \frac{b}{a} \)
- \( \alpha \beta = \frac{c}{a} \)

### Step 2: Rewrite the second equation
The second equation we need to analyze is:
\[ b^2c x^2 - ab^2x + a^3 = 0 \]

### Step 3: Normalize the second equation
To make it easier to work with, we can divide the entire equation by \( a^3 \):
\[ \frac{b^2c}{a^3} x^2 - \frac{b^2}{a} x + 1 = 0 \]

### Step 4: Identify coefficients
Let:
- \( A = \frac{b^2c}{a^3} \)
- \( B = -\frac{b^2}{a} \)
- \( C = 1 \)

### Step 5: Use Vieta's formulas for the new equation
The sum and product of the roots \( x_1 \) and \( x_2 \) of the quadratic equation \( Ax^2 + Bx + C = 0 \) are given by:
- \( x_1 + x_2 = -\frac{B}{A} = \frac{b^2/a}{b^2c/a^3} = \frac{a^3}{ac} = \frac{a^2}{c} \)
- \( x_1 x_2 = \frac{C}{A} = \frac{1}{b^2c/a^3} = \frac{a^3}{b^2c} \)

### Step 6: Find the roots
Now, we need to express the roots in terms of \( \alpha \) and \( \beta \). We know:
- \( \alpha + \beta = \frac{b}{a} \)
- \( \alpha \beta = \frac{c}{a} \)

Using these, we can express:
\[ x_1 = \frac{1}{\alpha^2 + \alpha \beta}, \quad x_2 = \frac{1}{\beta^2 + \alpha \beta} \]

### Step 7: Finalize the answer
Thus, the roots of the equation \( b^2c x^2 - ab^2x + a^3 = 0 \) are:
\[ \frac{1}{\alpha^2 + \alpha \beta}, \quad \frac{1}{\beta^2 + \alpha \beta} \]

Comparing with the options provided, we find that the correct answer is:
**Option b: \( \frac{1}{\alpha^2 + \alpha \beta}, \frac{1}{\beta^2 + \alpha \beta} \)**.