if the value of the determinant `|{:(a,,1,,1),(1,,b,,1),(1,,1,,c):}|` is positivie then
`(a,b,c lt 0)`

Options

A. `abc gt 1`
B. ` abc gt -8`
C. ` abc gt -8`
D. `abc gt -2`

Answer

B

Explanation

we have
`Delta =|{:(a,,1,,1),(1,,b,,1),(1,,1,,c):}| =abc -(a+b+c)+2`
`:. Delta gt 0 rArr abc+ 2 gt a+b+c`
`rArr abc+ 2 gt 3 (abc)^(1//3)`
`[ :' A.M gt G.M. rArr .(ab+c)/(3) gt (abc)^(1//3)]`
`rArr x^(3)+ 2 gt 3x , " where " x= (abc)^(1//3)`
`rArr x^(3)- 3x+2 gt 0 rArr (x-1)^(2) (x+2) gt 0`
`rArr x+2 gt 0 rArr x gt -2 rArr (abc)^(1//3) gt -2 rArr abc gt-8`