If the roots of the cubic `x ^(3) +ax ^(2) + bx +c=0` are three consecutive positive integers, then the value of `(a ^(2))/(b +1) =`

Options

A. `(a^(2))/(b+a)=3`
B. `(a^(2))/(b+a)=1`
C. `(b+1)/(a^(2))=3`
D. `(c^(2))/(a^(2))=1`

Answer

B, C, D

Explanation

`x^(2)+ax^(2)+bx+c=0larroverset(p-1)underset(p+1)p`
`a=-3p`
`b=p(p-1)+p(p+1)+(p-1)(p+1)`
`c=-p(p-1)(p+1)`
`(a^(2))/(b+a)=(9p^(2))/(p^(2)-cancel(p)+p^(2)+cancel(p)+p^(2)-cancel(1)+cancel(1))`
`=3`