Question

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

• A. $a=0,b=3$
• B. $a=b=0$
• C. $a=b=3$
• D. $a,b\text{are roots of}{x}^2+x+2=0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $a=b=0$
C $a=b=3$
D $a,b\text{are roots of}{x}^2+x+2=0$

Given equation is $x^3-a{x}^2+bx-1=0.$

If roots of the equation be $\alpha ,\beta ,\gamma ,$

then ${\alpha }^2+{\beta }^2+{\gamma }^2=(\alpha +\beta +\gamma {)}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )$

$=a^2-2b$

${\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2={(\alpha \beta +\beta \gamma +\gamma \alpha )}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )$

$=b^2-2a$

${\alpha }^2{\beta }^2{\gamma }^2=1$

So, the equation whose roots are ${\alpha }^2,{\beta }^2,{\gamma }^2$ is given by

$x^3-(a^2-2b)x^2+(b^2-2a)x-1=0$

It is identical to $x^3-a{x}^2+bx-1=0$

$\Rightarrow a^2-2b=a$ and $b^2-2a=b$

Eliminating $b,$ we get

$\frac{{(a^2-a)}^2}{4}-2a=\frac{a^2-a}{2}$

$\Rightarrow a\left(a\right(a-1{)}^2-8-2(a-1))=0$

$\Rightarrow a(a^3-2a^2-a-6)=0$

$\Rightarrow a(a-3)(a^2+a+2)=0$

$\Rightarrow a=0$ or $a=3$ or $a^2+a+2=0$

which gives $b=0$ or $b=3$ or $b^2+b+2=0.$

So, $a=b=0$ or $a=b=3$ or $a,b$ are roots of $x^2+x+2=0$