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If $\alpha$ and $\beta$ are the roots of the equation $a x^{2}+b x+c=0$, then the value of $\alpha^{3}+\beta^{3}$ is
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Verified Answer
The correct answer is:
$\frac{3 a b c-b^{3}}{a^{3}}$
Given : $\alpha \& \beta$ are roots of equation $a x^{2}+b x+c=0$
$\therefore \alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}} \& \alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}$
Now, $\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$
$\begin{array}{l}
\Rightarrow \alpha^{3}+\beta^{3}=\left(-\frac{b}{a}\right)^{3}-3 \frac{c}{a} \cdot\left(-\frac{b}{a}\right) \\
\Rightarrow \alpha^{3}+\beta^{3}=-\frac{b^{3}}{a^{3}}+\frac{3 b c}{a^{2}} \\
\Rightarrow \alpha^{3}+\beta^{3}=\frac{-b^{3}+3 a b c}{a^{3}}
\end{array}$
$\therefore \alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}} \& \alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}$
Now, $\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$
$\begin{array}{l}
\Rightarrow \alpha^{3}+\beta^{3}=\left(-\frac{b}{a}\right)^{3}-3 \frac{c}{a} \cdot\left(-\frac{b}{a}\right) \\
\Rightarrow \alpha^{3}+\beta^{3}=-\frac{b^{3}}{a^{3}}+\frac{3 b c}{a^{2}} \\
\Rightarrow \alpha^{3}+\beta^{3}=\frac{-b^{3}+3 a b c}{a^{3}}
\end{array}$
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