Question

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