Lower bound for $p_{n^3}-p_{(n-1)^3}$? The difference between two primes is at least $2$ so $p_{n^3}-p_{(n-1)^3} \geq 6n^2$. Is there any known sharper bound?

Looking at simple sieving,
there are at most 2 primes in
$[n, n+6)$,
8 primes in
$[n, n+30)$,
and so on for
the product of
for first $k$ primes,
the number being
$\prod_{i=1}^k (p_i-1)$.
Therefore
if
$r_k
=\prod_{i=1}^k \frac{p_i}{p_i-1}$,
$p_b-p_a
\ge (b-a)r_k
$.
Your case has
$b-a = 3n^2+3n+1$
and $k = 1$
with $r_1 = \frac{2}{1}
=2$.
With $k=2$,
$r_2
=\frac{2\cdot 3}{1\cdot 2}
=3$,
so we get
$9n^2$.
With $k=3$,
$r_3
=\frac{2\cdot 3\cdot 5}{1\cdot 2\cdot 4}
=\frac{15}{4}$,
so we get
$\frac{45}{4}n^2$.
In general,
the difference is
at least $3r_kn^2$.
By Merten's third theoren
(https://en.wikipedia.org/wiki/Mertens%27_theorems),
$\lim_{n \to \infty} 
\ln(n)\prod_{p \le n} (1-\frac1{p})
=e^{-\gamma}$,
where $\gamma$ is the Euler–Mascheroni constant.
Therefore
$\prod_{p \le n} \frac{p}{p-1}
\approx e^{\gamma}\ln(n)
$.
Therefore,
since
$p_k \approx k \ln(k)$,
$r_k
\approx e^{\gamma}\ln(k \ln k)
$.
To use $k$ primes,
you must have
$n^3 
\gt \prod_{i=1}^k p_i$.
Since
$ \prod_{p\le n} \ln p
\approx n$
(see https://en.wikipedia.org/wiki/Chebyshev_function),
this means that
$n^3
\gt e^{k\ln k+O(k \ln\ln k)}
$
so,
arguing non-rigorously,
$k\ln k+O(k \ln\ln k)
\le 3\ln(n)
$.
Choosing this
expression for
the max $k$,
we can get
$r_k
\approx e^{\gamma}\ln(3\ln(n))
$.
