If `alpha`, `beta`, `gamma` are the roots of the equation `9x^(3)-7x+6=0` then the equation `x^(3)+Ax^(2)+Bx+C=0` has roots `3alpha+2`, `3beta+2`, `3gamma+2`, where

Options

A. `A=6`
B. `B=-5`
C. `C=24`
D. `A+B+C=23`

Answer

C, D

Explanation

`(c,d)` Let `P=2alpha+2`
`implies alpha=(P-2)/(3)`
Since `9alpha^(3)-7alpha+6=0`
`implies (9(P-2)^(3))/(27)-(7)/(3)(P-2)+6=0`
`implies (1)/(3)(P^(3)-8-6P^(2)+12P)-(7)/(3)P+(14)/(3)+6=0`
`implies P^(3)-6P^(2)+12P-8-7P+14+18=0`
`impliesP^(3)-6P^(2)+5P+24=0`
so, the equation `x^(3)-6x^(2)+5x+24=0` has roots `3alpha+2`, `3beta+2`, `3gamma+2`.