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(B) Let the roots of the equation $x^3 - ax^2 + ax - 1 = 0$ be $\alpha, \beta, \gamma$. From Vieta's formulas,we have: $\alpha + \beta + \gamma = a$ $

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let the roots of the equation $x^3 - ax^2 + ax - 1 = 0$ be $\alpha, \beta, \gamma$. From Vieta's formulas,we have: $\alpha + \beta + \gamma = a$ $\alpha\beta + \beta\gamma + \gamma\alpha = a$ $\alpha\beta\gamma = 1$ The roots of the new equation are $\alpha^2, \beta^2, \gamma^2$. Since the new equation is identical to the original,the set of roots ${\alpha^2, \beta^2, \gamma^2}$ must be the same as ${\alpha, \beta, \gamma}$. Given $\alpha\beta\gamma = 1$,we consider the case where the roots are permuted. If $\alpha^2 = \alpha, \beta^2 = \beta, \gamma^2 = \gamma$,then $\alpha, \beta, \gamma \in {0, 1}$. Since the product is $1$,all roots must be $1$. Then $x^3 - ax^2 + ax - 1 = (x-1)^3 = x^3 - 3x^2 + 3x - 1$. Comparing coefficients,$a = 3$. Checking: if $a=3$,the roots are $1, 1, 1$. Their squares are $1, 1, 1$,which are the same roots. Thus,$a = 3$.

Column-$I$	Column-$II$
$A$. $\alpha+\beta+\gamma$	$(1)$ $-\frac{43}{4}$
$B$. $\alpha^2+\beta^2+\gamma^2$	$(2)$ $-\frac{7}{8}$
$C$. $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$	$(3)$ $86$
$D$. $\frac{\alpha}{\beta\gamma}+\frac{\beta}{\gamma\alpha}+\frac{\gamma}{\alpha\beta}$	$(4)$ $0$
	$(5)$ $10$

Let $a$ be a non-zero real number. If the equation whose roots are the squares of the roots of the cubic equation $x^3 - ax^2 + ax - 1 = 0$ is identical to the original cubic equation,then $a =$

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