Question

If $\alpha \, and \, \beta$ are the roots of the equation ax+bx+c=0, then the value of $\alpha^{3} \, + \, \beta^{3}$ is

• A. $\frac{3abc \, + \, b^{3}}{a^3}$
• B. $\frac{a^3 \, + \, b^3}{3abc}$
• C. $\frac{3abc \, - \, b^3}{a^3}$
• D. $\frac{-(3abc \, + \, b^3)}{a^3}$

Answer 1

The Correct Option is C

To find the value of $\alpha^{3} + \beta^{3}$ when $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$, we can use the properties of roots for quadratic equations.

1. By Vieta's formulas, for a quadratic equation $ax^2 + bx + c = 0$, the sum and product of the roots $\alpha$ and $\beta$ are given by:
   • $\alpha + \beta = -\frac{b}{a}$
   • $\alpha \beta = \frac{c}{a}$
2. Let's derive the expression for $\alpha^3 + \beta^3$ using the identity:
   $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)$
3. We can express $\alpha^2 - \alpha \beta + \beta^2$ as:
   $\alpha^2 - \alpha \beta + \beta^2 = (\alpha + \beta)^2 - 3\alpha \beta$
4. Substitute $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$ into the equation:
   $\alpha^2 - \alpha \beta + \beta^2 = \left(-\frac{b}{a}\right)^2 - 3 \cdot \frac{c}{a}$
   $\alpha^2 - \alpha \beta + \beta^2 = \frac{b^2}{a^2} - \frac{3c}{a}$
5. So, the expression for $\alpha^3 + \beta^3$ becomes:
   $\alpha^3 + \beta^3 = \left(-\frac{b}{a}\right) \left(\frac{b^2}{a^2} - \frac{3c}{a}\right)$
   $\alpha^3 + \beta^3 = -\frac{b}{a} \cdot \frac{b^2}{a^2} + \frac{3b}{a} \cdot \frac{c}{a}$
   $\alpha^3 + \beta^3 = -\frac{b^3}{a^3} + \frac{3bc}{a^2}$
   $\alpha^3 + \beta^3 = \frac{3abc - b^3}{a^3}$

The correct option is the one which matches our derived expression:

• $\frac{3abc - b^3}{a^3}$ - This is the derived formula we calculated.

Therefore, the correct answer is $\frac{3abc - b^3}{a^3}$.