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If $\alpha, \beta, \gamma$ are roots of $x^3-5 x+4=0$, then $\left(\alpha^3+\beta^3+\gamma^3\right)^2$ is equal to
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The correct answer is:
144
Given, the roots of
$$
\begin{array}{rlrl}
& & x^3-5 x+4 & =0 \text { are } \alpha \beta \text { and } \gamma \\
\therefore & \alpha+\beta+\gamma & =0
\end{array}
$$
Since,
$$
\alpha \beta+\beta \gamma+\gamma \alpha=5 \text { and } \alpha \beta \gamma=-4
$$
$\therefore$
$\alpha+\beta+\gamma=0$
$$
\begin{aligned}
\alpha^3+\beta^3+\gamma^3 & =3 \alpha \beta \gamma \\
& =3 \times(-4)=-12
\end{aligned}
$$
Hence, $\left(\alpha^3+\beta^3+\gamma^3\right)^2=(-12)^2=144$
$$
\begin{array}{rlrl}
& & x^3-5 x+4 & =0 \text { are } \alpha \beta \text { and } \gamma \\
\therefore & \alpha+\beta+\gamma & =0
\end{array}
$$
Since,
$$
\alpha \beta+\beta \gamma+\gamma \alpha=5 \text { and } \alpha \beta \gamma=-4
$$
$\therefore$
$\alpha+\beta+\gamma=0$
$$
\begin{aligned}
\alpha^3+\beta^3+\gamma^3 & =3 \alpha \beta \gamma \\
& =3 \times(-4)=-12
\end{aligned}
$$
Hence, $\left(\alpha^3+\beta^3+\gamma^3\right)^2=(-12)^2=144$
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