If ` alpha and beta` are the roots of `ax^(2)+bx+c=0`, then the value of the expression `(a alpha +b)^(-3)+ (a beta +b)^(-3)` is equal to

Options

A. `(a^(3)-3abc)/(b^(3)c^(3))`
B. `(b^(3)-3abc)/(c^(3)a^(3))`
C. `(c^(3)-3abc)/(alpha^(3)b^(3))`
D. none

Answer

To solve the problem, we need to find the value of the expression \((a \alpha + b)^{-3} + (a \beta + b)^{-3}\), where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(ax^2 + bx + c = 0\).

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of the quadratic equation \(ax^2 + bx + c = 0\) are given by:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{(sum of roots)} \]
\[ \alpha \beta = \frac{c}{a} \quad \text{(product of roots)} \]

2. **Rewrite the Expression**:
We need to evaluate:
\[ (a \alpha + b)^{-3} + (a \beta + b)^{-3} \]
This can be rewritten as:
\[ \frac{1}{(a \alpha + b)^3} + \frac{1}{(a \beta + b)^3} \]

3. **Find a Common Denominator**:
The common denominator for the two fractions is:
\[ (a \alpha + b)^3 (a \beta + b)^3 \]
Thus, we can express the sum as:
\[ \frac{(a \beta + b)^3 + (a \alpha + b)^3}{(a \alpha + b)^3 (a \beta + b)^3} \]

4. **Use the Sum of Cubes Formula**:
The sum of cubes can be expressed using the formula:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
Let \(x = a \alpha + b\) and \(y = a \beta + b\). Then:
\[ x + y = (a \alpha + b) + (a \beta + b) = a(\alpha + \beta) + 2b = a\left(-\frac{b}{a}\right) + 2b = -b + 2b = b \]
Now, we need to calculate \(x^2 - xy + y^2\):
\[ x^2 + y^2 = (a \alpha + b)^2 + (a \beta + b)^2 \]
Expanding these:
\[ (a \alpha + b)^2 = a^2 \alpha^2 + 2ab \alpha + b^2 \]
\[ (a \beta + b)^2 = a^2 \beta^2 + 2ab \beta + b^2 \]
Adding these:
\[ x^2 + y^2 = a^2 (\alpha^2 + \beta^2) + 2ab (\alpha + \beta) + 2b^2 \]
Using \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\):
\[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} \]
Thus:
\[ x^2 + y^2 = a^2\left(\frac{b^2}{a^2} - \frac{2c}{a}\right) + 2ab\left(-\frac{b}{a}\right) + 2b^2 \]
Simplifying gives:
\[ b^2 - 2ac - 2b^2 + 2b^2 = -2ac + b^2 \]
Now, we also need \(xy = (a \alpha + b)(a \beta + b) = a^2 \alpha \beta + ab(\alpha + \beta) + b^2 = a c - b^2 + b^2 = ac\).

5. **Combine Everything**:
Now substituting back into the sum of cubes:
\[ x^3 + y^3 = b\left((-2ac + b^2) - ac\right) = b(-3ac + b^2) \]
Thus, the expression becomes:
\[ \frac{b(-3ac + b^2)}{(a \alpha + b)^3 (a \beta + b)^3} \]

6. **Final Expression**:
The denominator simplifies to:
\[ (a^2 c^2) \quad \text{(since both roots contribute)} \]
Therefore, the final value of the expression is:
\[ \frac{b(-3ac + b^2)}{(ac)^3} \]

### Final Answer:
The value of the expression is:
\[ \frac{b^3 - 3abc}{(ac)^3} \]