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

Answer

2,3,4

Explanation

Given equation is `x^(3) - ax^(3) + bx - 1 = 0 ` . If roots of the equationo
be ` alpha, beta , gamma`, then
`alpha^(2) + beta^(2) + gamma^(2) = (alpha + beta + gamma)^(2) - 2 (alpha beta + beta gamma + gamma alpha)`
` = alpha ^(2) - 2a `
`alpha^(2)beta^(2) + beta^(2) gamma^(2)+ gamma^(2)alpha ^(2) = (alpha beta + beta gamma+ gammaalpha )^(2) - 2 alpha beta gamma (alpha + beta + gamma )`
` b^(2) - 2a `
` alpha ^(2) beta^(2) gamma^(2) = 1`
So, the equations whose roots are ` alpha ^(2) , beta^(2) , gamma^(2)` is given by
`x^(3) - ax^(2) + bx - 1= 0 `
`rArr a^(2) - 2b = a and b^(2) - 2a = b`
Eliminatng b, we get
`((a^(2) - a)^(2))/(4) - 2a = (a^(2) - a)/(2)`
or ` a{a (a - 1)^(2) - 8 - 2a (a - 1)} = 0 `
or ` a (a^(3) - 2a ^(2) + a + 2) = 0`
or ` a(a - 3) (a^(2) + a + 2) = 0`
`rArr a = 0 or a = 3 or a^(2) + a + 2 = 0`
Which gives b = 0 or b = 3 `b^(2) + b+ 2 = 0 `.So a = b - 0 or
a = b =3 or a a, b are roots of ` x^(2) + x + 2 = 0`