If `alpha and beta` are the roots of the equation `ax ^(2) + bx + c=0,a,b, c in R , alpha ne 0` then which is (are) correct:

Options

A. `alpha ^(2) +beta ^(2) =(b ^(2) -2ac)/(a ^(2))`
B. `(1)/(alpha^(2))+(1)/(beta^(2))=(b^(2) -2ac)/(c ^(2))`
C. `(1)/(alpha^(3))+(1)/(beta^(3))=(abc -b^(3))/(c ^(3))`
D. `alpha beta (alpha+beta ) =(-bc)/(alpha ^(2))`

Answer

To solve the problem, we need to analyze the given quadratic equation \( ax^2 + bx + c = 0 \) and its roots \( \alpha \) and \( \beta \). We will use Vieta's formulas, which relate the coefficients of the polynomial to sums and products of its roots.

### Step 1: Identify the relationships between the roots and coefficients
From Vieta's formulas, we have:
1. \( \alpha + \beta = -\frac{b}{a} \) (sum of the roots)
2. \( \alpha \beta = \frac{c}{a} \) (product of the roots)

### Step 2: Calculate \( \alpha^2 + \beta^2 \)
We can express \( \alpha^2 + \beta^2 \) using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values from Vieta's formulas:
\[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) \]
\[ = \frac{b^2}{a^2} - \frac{2c}{a} \]
\[ = \frac{b^2 - 2ac}{a^2} \]

### Step 3: Calculate \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \)
We can write:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]
Using the results from Step 2 and the product from Step 1:
\[ = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = \frac{\frac{b^2 - 2ac}{a^2}}{\left(\frac{c}{a}\right)^2} \]
\[ = \frac{b^2 - 2ac}{a^2} \cdot \frac{a^2}{c^2} = \frac{b^2 - 2ac}{c^2} \]

### Step 4: Calculate \( \frac{1}{\alpha^3} + \frac{1}{\beta^3} \)
Using the identity:
\[ \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\alpha^3 + \beta^3}{\alpha^3 \beta^3} \]
We know:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha \beta) \]
Substituting the known values:
\[ = \left(-\frac{b}{a}\right)\left(\frac{b^2 - 2ac}{a^2} - \frac{c}{a}\right) \]
\[ = \left(-\frac{b}{a}\right)\left(\frac{b^2 - 2ac - ac}{a^2}\right) = \left(-\frac{b}{a}\right)\left(\frac{b^2 - 3ac}{a^2}\right) \]
Thus:
\[ \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{-\frac{b(b^2 - 3ac)}{a^3}}{\left(\frac{c}{a}\right)^3} = \frac{-b(b^2 - 3ac)}{c^3} \]

### Step 5: Calculate \( \alpha \beta (\alpha + \beta) \)
We can express this as:
\[ \alpha \beta (\alpha + \beta) = \left(\frac{c}{a}\right)\left(-\frac{b}{a}\right) = -\frac{bc}{a^2} \]

### Conclusion
After calculating the values for each expression, we conclude that all derived expressions are correct based on the relationships established by Vieta's formulas. Therefore, all options provided in the question are correct.