Statement-I     The zeroes $x^3-3p{x}^2+qx-r$  are in A.P., then $2p^3=pq-r$ 

$\mathbf{Statement}\mathbf{-}\mathbf{II}\mathbf{ }\mathbf{ } \mathrm{If} {\mathrm{ax}}^3-\mathrm{bx}-\mathrm{c} \mathrm{is} \mathrm{divisible} \mathrm{by} {\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}, \mathrm{then} \mathrm{ab}=1$

Let $f\left(x\right)=x^{3^{-}}-3p{x}^{2^{+}}+q{x}^{-}-r$ .

Given that  the zeros are in A.P

$\Rightarrow a-d+a+a+d=3p\Rightarrow 3a=3p\Rightarrow a>p$

$\therefore f\left(p\right)=p^3-3p^3+pq-r=-2p^3+pq-r=0$

$\Rightarrow 2p^3=r-pq$

$\therefore$  statement (A) is FALSE

Statement (B)

$\Rightarrow a{b}^2-b-ac=0$

$c\left(ab-1\right)=b$

$\therefore c\ne 0\Rightarrow ab=1$

$\therefore$  statement (B) is TRUE