Let `Delta = |{:(-bc,,b^(2)+bc,,c^(2)+bc),(a^(2)+ac,,-ac,,c^(2)+ac),(a^(2)+ab,,b^(2)+ab,,-ab):}|` and the equation
`px^(3) +qx^(2) +rx+s=0` has roots a,b,c where `a,b,c in R^(+)`
the value of `Delta ` is

Options

A. `r^(2)//p^(2)`
B. `r^(3)//p^(3)`
C. `-s//p`
D. none of these

Answer

B

Explanation

Multiplying `R_(1) ,R_(2),R_(3) " by " a,b,c ` respectively ,and then taking a,b,c common from` C_(1),C_(2)" and " C_(3)` we get
`Delta = |{:(-bc,,ab+ac,,ac+ab),(ab+bc,,-ac,,bc+ab),(ac+bc,,bc+ac,,-ab):}|`
Now using `C_(2) to C_(2)-C_(1)" and " C_(3) to C_(3) -C_(1) ` and then taking (ab+bc+ca) common from `C_(2) " and " C_(3)` we get
`Delta =|{:(-bc,,1,,1),(ab+bc,,-1,,0),(ac+bc,,0,,-1):}|xx (ab +bc +ca)^(2)`
Now applying `R_(2) to R_(2) +R_(1)` we get
`Delta = |{:(-bc,,1,,1),(ab,,0,,1),(ac+bc,,0,,-1):}| (ab+bc+ca)^(2)`
Expanding along `c_(2)` we get
`Delta =(ab+bc+ca)^(2)[ac+bc+ca)^(2)`
`=(ab+bc+ca)^(2)`
`=(r//p)^(3) =r^(3)//p^(3)`
Now given a,b,c are all positive then
`A.M ge G.M.`
`rArr (ab+bc+ac)/(3) ge (abxx bcxx ac)^(1//3)`
`" or " (ab+bc+ac)^(3) ge 27a^(2)b^(2)c^(2)`
`" or " (ab+bc+ca)^(3) ge 27(s^(2)//p^(2))`
if `Delta =27` then ab+bc+ca =3 and given that `a^(2) +b^(2)+c^(2)=3`
From `(a+b+c)^(2) =a^(2) +b^(2)+c^(2) +2 (ab+bc+ca)` we have
`a+b+c = ne 3`
`rArr a+b+c =3`
`rArr 3p+q=0`