Let `alpha,beta,gamma` be the roots of `(x-a) (x-b) (x-c) = d, d != 0`, then the roots of the equation `(x-alpha)(x-beta)(x-gamma) + d =0` are `:`

Options

A. a,b,d
B. b,c,d
C. a,b,c
D. `a+d,b+d,c+d`

Answer

C

Explanation

Since `alpha, beta` and `gamma` are the roots of
`(x-a)(x-b)(x-c)=d`
`implies(x-a)(x-b)(x-c)-d(x-alpha)(x-beta)(x-gamma)`
`implies(x-alpha)(x-beta)(x-gamma)+d=(x-a)(x-b)(x-c)`
`impliesa,b`and `c` are the roots of
`(x-alpha)(x-beta)(x-gamma)+d=0`