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If one root of the quadratic equation $a x^2+b x+c=0$ is equal to the $n$th power of the other, then $\left(a c^n\right)^{1 / n+1}+\left(a^n c\right)^{1 / n+1}=$
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The correct answer is:
$-b$
Let $\alpha$ and $\alpha^n$ are the roots of the equation
$$
\begin{aligned}
\alpha+ & \alpha^n=-\frac{b}{a} \text { and } \alpha^{n+1}=\frac{c}{a} \\
& =\left(a c^n\right)^{\frac{1}{n+1}}+\left(a^n c\right)^{\frac{1}{n+1}} \\
& =\left[a\left(a \alpha^{n+1}\right)^n\right]^{\frac{1}{n+1}}+\left(a^n \times a \alpha^{n+1}\right)^{\frac{1}{n+1}} \\
& =\left[a^{n+1} \alpha^{n(n+1)}\right]^{\frac{1}{n+1}}+\left(a^{n+1} \alpha^{n+1}\right)^{\frac{1}{n+1}} \\
& =a \alpha^n+a \alpha=a\left(\alpha^n+\alpha\right) \\
& a\left(-\frac{b}{a}\right)=-b
\end{aligned}
$$
$$
\begin{aligned}
\alpha+ & \alpha^n=-\frac{b}{a} \text { and } \alpha^{n+1}=\frac{c}{a} \\
& =\left(a c^n\right)^{\frac{1}{n+1}}+\left(a^n c\right)^{\frac{1}{n+1}} \\
& =\left[a\left(a \alpha^{n+1}\right)^n\right]^{\frac{1}{n+1}}+\left(a^n \times a \alpha^{n+1}\right)^{\frac{1}{n+1}} \\
& =\left[a^{n+1} \alpha^{n(n+1)}\right]^{\frac{1}{n+1}}+\left(a^{n+1} \alpha^{n+1}\right)^{\frac{1}{n+1}} \\
& =a \alpha^n+a \alpha=a\left(\alpha^n+\alpha\right) \\
& a\left(-\frac{b}{a}\right)=-b
\end{aligned}
$$
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