Question

If the cubic equation $x^{3}-a x^{2}+a x-1=0$ is identical with the cubic equation whose roots are the squares of the roots of the given cubic equation, then the non-zero real value of ${ }^{'} a^{'}$ is

• A. $\frac{1}{2}$
• B. 2
• C. 3
• D. $\frac{7}{2}$

Answer 1

Answer: (C)

Let $\alpha, \beta, \gamma$ are roots of equation
$x^{3}-a x^{2}+a x-1=0\,\,\,...(i)$
$\therefore \alpha+\beta+\gamma =a $
$\alpha \beta+\beta \gamma+\alpha \gamma =a$
$\alpha \beta \gamma =-1$
Cubic equation whose roots are $\alpha^{2}, \beta^{2}, \gamma^{2}$ is $x^{2}-\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right) x^{2}+\left(\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\alpha^{2} \gamma^{2}\right) x$
$-\alpha^{2} \beta^{2} \gamma^{2}=0 \ldots$ (ii)
Eqs. (i) and (ii) are identical.
$\therefore \frac{a}{\alpha^{2}+\beta^{2}+\gamma^{2}}=\frac{a}{\alpha^{2} \beta^{2}+\beta^{2} \gamma^{2}+\gamma^{2} \alpha^{2}}=\frac{1}{\alpha^{2} \beta^{2} \gamma^{2}}$
$a=\alpha^{2}+\beta^{2}+\gamma^{2} \,\,\,[\alpha \beta \gamma=-1]$
$a=(\alpha+\beta+\gamma)^{2}-2(\alpha \beta+\beta \gamma+\gamma \alpha)$
$a=a^{2}-2 a \Rightarrow a^{2}=3 a$
$\Rightarrow a=3\,\,\, [\because a$ is non-zero real $]$