If `alpha,beta,gamma` are the roots of `a x^3+b x^2+cx+d=0` and `|[alpha,beta,gamma],[beta,gamma,alpha],[gamma,alpha,beta]|=0, alpha!=beta!=gamma` then find the equation whose roots are `alpha+beta-gamma,beta+gamma-alpha`, and `gamma+alpha-beta`.

Answer

`ax^(3)-2bx^(2)+4cx-8d=0`

Explanation

`|{:(alpha,,beta,,gamma),(beta,,gamma,,alpha),(gamma,,alpha,,beta):}|=(alpha+beta+gamma).(1)/(2)[(alpha-beta)^(2)+(beta-gamma)^(2)+(gamma-alpha)^(2)]=0`
since `alpha ne beta ne gamma` we have
`alpha +beta +gamma=0`
`:. Alpha +beta-gamma =-2gamma`
`beta+gamma-alpha=-2alpha`
`gamma+alpha-beta=-2beta`
`" Let " y=-2x`
`" or " x=(-y)/(2)` Therefore the required equation is
`alpha (-(y)/(3))^(3)+b(-(y)/(2))^(2)+c(-(y)/(2))+d=0`
`rArr (-a)/(8) y^(3)+(b)/(4)y^(2)-(c)/(l2)y+d=0`
`rArr ay^(3)-2by^(2)+4cy-8d=0` lt brgt `" or " ax^(3) -2bx^(2)+4cx -8d=0`