If the equation whose roots are the squares of the roots of the cubic `x^3-a x^2+b x-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 ,` are roots of`x^2+x+2=0`

Options

A. `a=b=0`
B. `a=0,b=3`
C. `a=b=3`
D. `a,b` are roots of `x^(2)+x+2=0`

Answer

A, C, D

Explanation

We have `x^(3)-ax^(2)+bx-1=0`……….i
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`
and `alpha^(2) beta^(2) gamma^(2)=1`
Therefore, the equation whose roots are`alpha^(2),beta^(2)` and `gamma^(2)` is
`x^(3)-(a^(2)-2b)x^(2)+(b^(2)-2a)x-1=0`...........ii
`a^(2)-2b=a` and `b^(2)-2a=b`
Eliminating `b` we have
`((a^(2)-a)^(2))/4-2a=(a^(2)-a)/2`
`impliesa{a(a-1)^(2)-8-2(a-1)}=0`
`impliesa(a^(3)-2a^(2)-a-6)=0`
`impliesa(a-3)(a^(2)+a+2)=0`
`impliesa=0` or `a=3` or `a^(2)=a+2=0`
`impliesb=0` or `b=3`
or `b^(2)+b+2=0`
`:.a=b=0`
or `a=b=3`
or a and b are roots of `x^(2)+x+2=0`