Correct Answer - `x=-1,(1)/(2),(2)/(3),-(2)/(3)`
`|{:(x^(2)-1,,x^(2)+2x+1,,2x^(2)+3x+1),(2x^(2)+x-1,,2x^(2)+5x-3,,4x^(2)+4x-3),((3x-2)(2x+1),,6x^(2)-7x+2,,12x^(2)-5x-2):}|=0`
`" or " |{:((x-1)(x+1),,(x+1)^(2),,(x+1)(2x+1)),((2x-1)(x+1),,(2x-1)(x+3),,(2x-1)(2x+3)),((3x-2)(2x+1),,(3x-2)(2x+1),,(3x-2)(4x+1)):}|=0`
`" or " (x+1)(2x-1)(3x-2) |{:(x-1,,x+1,,2x+1),(x+1,,x+3,,2x+3),(2x+1,,2x-1,,4x+1):}|=0`
`C_(2) to C_(2)-C_(1) " and "C_(3) to C_(3) -2C_(1)`
`rArr (x+1) (2x-1) (3x-2) |{:(x-1,,2,,3),(x+1,,2,,1),(2x+1,,-2,,-1):}|=0`
Applying `C_(2)to C_(2)-2C_(3)` we get
`(x+1) (2x-1) (3x-2) |{:(x-1,,-4,,3),(x+1,,0,,1),(2x+1,,0,,-1):}|=0`
`" or " 4(x+1)(2x-1)(3x-2)(3x+2)=0`
`" or " x=-1, (1)/(2),(2)/(3),(2)/(3)`
