i explain with another method,first by ingham theorem we have $M<q<M+AM^{5/8}$,by substituting $N$ instead to $M$,and $q$ to $q_1$ ,$N<q_1<N+AN6{5/8}$,now by again substituting $N+AN^{5/8}$ instead to $M$,SO $N+AN^{5/8}<q_2<N+AN^{5/8}+(N+AN^{5/8})^{5/8}$, and so on,we can reach to cube

thank you but i need complete solution

in above answer first we reach to $N< P_{n+1}< N+AN^[5/8}$,so we have at least one prime between this intervals,second,we use this method to reach to other intervals,by induction,notice that $q_2$ may is not equal to $p_{n+2} i.e $q_2=p_{n+s_1} in which $s_1\ge 2$

which part of above answer is not clear?