Question:
If $\alpha \, and \, \beta$ are the roots of the equation ax+bx+c=0, then the value of $\alpha^{3} \, + \, \beta^{3}$ is
If $\alpha \, and \, \beta$ are the roots of the equation ax+bx+c=0, then the value of $\alpha^{3} \, + \, \beta^{3}$ is
Updated On: Jun 23, 2023
- $\frac{3abc \, + \, b^{3}}{a^3}$
- $\frac{a^3 \, + \, b^3}{3abc}$
- $\frac{3abc \, - \, b^3}{a^3}$
- $\frac{-(3abc \, + \, b^3)}{a^3}$
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The Correct Option is C
Solution and Explanation
Given : $\alpha$ & $\beta$ are roots of equation
ax + bx + c = 0
$\therefore \, \, \alpha \, + \beta \, = -\frac{b}{a} \, \& \, \alpha \beta \, = \, \frac{c}{a}$
Now, $\alpha^3 + \beta^3 \, =(\alpha \, + \beta)^3 - \, 3\alpha\beta(\alpha+\beta)$
$\Rightarrow \, \alpha^3+\beta^3 = \bigg(-\frac{b}{a}\bigg)^3 \, -3 \frac{c}{a} \bigg(-\frac{b}{a}\bigg)$
$\Rightarrow \, \, \alpha^3 + \beta^3 \, = \, -\frac{b^2}{a^3} + \frac{3bc}{a^2}$
$\Rightarrow \, \, \, \alpha^3 + \beta^3 \, = \, \frac{-b^3 \, + \, 3abc}{a^3}$
ax + bx + c = 0
$\therefore \, \, \alpha \, + \beta \, = -\frac{b}{a} \, \& \, \alpha \beta \, = \, \frac{c}{a}$
Now, $\alpha^3 + \beta^3 \, =(\alpha \, + \beta)^3 - \, 3\alpha\beta(\alpha+\beta)$
$\Rightarrow \, \alpha^3+\beta^3 = \bigg(-\frac{b}{a}\bigg)^3 \, -3 \frac{c}{a} \bigg(-\frac{b}{a}\bigg)$
$\Rightarrow \, \, \alpha^3 + \beta^3 \, = \, -\frac{b^2}{a^3} + \frac{3bc}{a^2}$
$\Rightarrow \, \, \, \alpha^3 + \beta^3 \, = \, \frac{-b^3 \, + \, 3abc}{a^3}$
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There are basically four methods of solving quadratic equations. They are:
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