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Let $\alpha \neq 1$ be a real root of the equation $x^3-a x^2+a x-1=0$, where $a \neq-1$ is a rea number. Then, a root of this equation, among the following, is
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The correct answer is:
$\frac{1}{\alpha}$
Equation $x^3-a x^2+a x-1=0$ and $\alpha \neq 1$, $a \neq-1$.
we put, $\left(x=\frac{1}{y}\right)$ in given equation
Then, $\left(\frac{1}{y}\right)^3-a\left(\frac{1}{y}\right)^2+a\left(\frac{1}{y}\right)-1=0$
$\frac{1}{y^3}-\frac{a^2}{y^2}+\frac{a}{y}-1=0$
$1-a^2 y+a y^2-y^3=0$
$\Rightarrow \quad y^3-a y^2+a^2 y-1=0$
Since, the reduced equation is same as original equation by replacing $\left(x=\frac{1}{y}\right) i e$, reciprocal root of the given equation.
Hence, $\left(x=\frac{1}{\alpha}\right)$ is a root of the given equation
we put, $\left(x=\frac{1}{y}\right)$ in given equation
Then, $\left(\frac{1}{y}\right)^3-a\left(\frac{1}{y}\right)^2+a\left(\frac{1}{y}\right)-1=0$
$\frac{1}{y^3}-\frac{a^2}{y^2}+\frac{a}{y}-1=0$
$1-a^2 y+a y^2-y^3=0$
$\Rightarrow \quad y^3-a y^2+a^2 y-1=0$
Since, the reduced equation is same as original equation by replacing $\left(x=\frac{1}{y}\right) i e$, reciprocal root of the given equation.
Hence, $\left(x=\frac{1}{\alpha}\right)$ is a root of the given equation
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