Prime

$(p - 1)^3 \equiv -1 \pmod p$ while $134p - 14 \equiv -14 \pmod p$ . Since LHS equal to RHS , $-1 \equiv -14 \pmod p$ . So , $13$ is divisible by $p$ . The only possible value for $p$ is $13$ . Check $p = 13$ onto both sides yields equality . Therefore , the solution is $p = 13$

rearrange the equation by taking 2 as common in RHS,now given that p is prime and whole cube of LHS is equal to RHS so whole cube of (p-1) must have two as one of its factor,now in RHS by using congruences we can conclude that 3p-7 is congruent to 0 mod 4(as 2(67p-7) must be divisible by 8 so 67p-7 must be divisible by 4) so we can write 3p-4r=7,its a linear diophantine eqn (r is an integer).on solving p=5+4k(k is an integer) and k = 2 gives p=13.

p^3 - 3p^2 + 3p - 1 = 134p - 14

p^3 - 3p^2 - 131p = -13

p(p^2 -3p -131) = -13

p has to be prime, so the only possible solution is 13.