A number n has 3 divisors. How many divisors does $n^3$ have?
n ==9
(1, 3, 9)>>Total = 3 divisors
9^3==729
(1, 3, 9, 27, 81, 243, 729)>>Total = 7 divisors
n has 3 divisors, these muct be 1, itself and one other, which means that n is a perfect square (of a prime number)
so
n=a^2 where a is prime
n^3=(a^2)^3 = a^6
This will have 6+1=7 factors (divisors)
