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

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

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$\frac{1}{\alpha}$

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(C) Given equation: $x^3-ax^2+ax-1=0$. We can factorize the equation as: $(x^3-1) - ax(x-1) = 0$. $(x-1)(x^2+x+1) - ax(x-1) = 0$. $(x-1)(x^2+x+1-ax) = 0$. Since $\alpha 
eq 1$ is a root,$\alpha$ must satisfy $x^2+(1-a)x+1=0$. Thus,$\alpha^2+(1-a)\alpha+1=0$. If $x = \frac{1}{\alpha}$ is a root,then $(\frac{1}{\alpha})^3 - a(\frac{1}{\alpha})^2 + a(\frac{1}{\alpha}) - 1 = 0$. Multiplying by $\alpha^3$,we get $1 - a\alpha + a\alpha^2 - \alpha^3 = 0$,which is $-(x^3-ax^2+ax-1) = 0$ evaluated at $x=\alpha$. Since $\alpha$ is a root,$\frac{1}{\alpha}$ is also a root of the equation.

Let $\alpha \neq 1$ be a real root of the equation $x^3-ax^2+ax-1=0$,where $a \neq -1$ is a real number. Then,a root of this equation,among the following,is

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