Discussion Forum

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 'a'  is 

Answer:

Explanation:

Let $\alpha,\beta ,\gamma$  are roots of equation

    $x^{3}-ax^{2}+ax-1$=0     ..........(i)

 $\therefore$          $\alpha$+$\beta$+$\gamma$=a

    $\alpha\beta +\beta\gamma+\alpha\gamma=a$

        $\alpha\beta \gamma=-1$

Cubic equation whose  roots  $\alpha^{2}$, $\beta^{2}$, $\gamma ^{2}$ is

 $x^{2}-(\alpha^{2}+\beta^{2}+ \gamma^{2})x^{2}+(\alpha^{2} \beta^{2}+\beta^{2}\gamma^{2}+\alpha^{2} \gamma^{2})x-\alpha ^{2}\beta^{2} \gamma ^{2}=0$     ........(ii)

 Equi .(i)  and (ii)  are identical.

$\therefore\frac{a}{\alpha^{2}+\beta^{2}+ \gamma^{2}}=\frac{a}{\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+ \alpha^{2}\gamma^{2}}=\frac{1}{\alpha^{2}\beta^{2} \gamma^{2}}$

 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 ]