Let alpha neq 1 be a real root of the equatio | vedclass

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(C) Given the equation $x^3-a x^2+a x-1=0$. We can factorize the equation as follows: $(x^3-1) - a x(x-1) = 0$ $(x-1)(x^2+x+1) - a x(x-1) = 0$ $(x-1)(

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$\frac{1}{\alpha}$

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(C) Given the equation $x^3-a x^2+a x-1=0$. We can factorize the equation as follows: $(x^3-1) - a x(x-1) = 0$ $(x-1)(x^2+x+1) - a x(x-1) = 0$ $(x-1)(x^2+x+1-a x) = 0$ $(x-1)(x^2+(1-a)x+1) = 0$. Since $\alpha 
eq 1$ is a root,$\alpha$ must satisfy the quadratic equation $x^2+(1-a)x+1=0$. Thus,$\alpha^2+(1-a)\alpha+1=0$. Dividing by $\alpha$ (since $\alpha 
eq 0$ as $0$ is not a root),we get $\alpha + (1-a) + \frac{1}{\alpha} = 0$,which implies $\frac{1}{\alpha} = a - 1 - \alpha$. Alternatively,note that if $x$ is a root,then $\frac{1}{x}$ is also a root because the equation is reciprocal. Substituting $x = \frac{1}{y}$ into the original equation: $(\frac{1}{y})^3 - a(\frac{1}{y})^2 + a(\frac{1}{y}) - 1 = 0$ $1 - ay + ay^2 - y^3 = 0$ $y^3 - ay^2 + ay - 1 = 0$. This is the same equation as the original. Therefore,if $\alpha$ is a root,then $\frac{1}{\alpha}$ is also a root.

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 real number. Then,a root of this equation,among the following,is

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