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If `alpha`,`beta``gamma` are the roots of the equation `x^3+px^2+qx+r=0`, then `(1-alpha^2)(1-beta^2)(1-gamma^2)` is equal to
(a)   `(1+q)^2-(p-r)^2`   (b)   `(1+q)^2+(p+r)^2`
(c)   `(1-q)^2+(p-r)^2`   (d)   none of these

A

`(1+q)^(2))-(p+r)^(2)`

B

`(1+q)^(2)+(p+r)^(2)`

C

`(1-q)^(2)+(p-r)^(2)`

D

`(1+r)^(2)-(p+q)^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A

`x^(3)+px^(2)+qx+r=0`
`x^(3)+px^(2)+qx+r=(x-alpha)(alpha-beta)(x-gamma)=1+p+q+r`…………1
pur `x=-1`
`implies-(1+alpha)(1+beta)(1+gamma)=-1+p-q+r`……..2
Multiplying (1) & (2)
`(1-alpha^(2))(1-beta^(2))(1-gamma^(2)))={(1+q)+(p+r))((1+q)-(p+r)}`
`=(1+q)^(2)-(p+r)^(2)`
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