Problems need HardWork #4

$P\times P \times P=P \times P \times P \times1$

There are $\boxed{4}$ distinct factor. It can't be others because $P$ is a prime, and $1$ has only 1 divisor.

Given that p is prime, the factors of p^3 are: 1, p, p^2, and p^3

$p^3 = \begin{cases} 1 \times p^3 \\ p \times p^2 \end{cases}$ . Therefore, there are $\boxed{4}$ factors: $1$ , $p$ , $p^2$ and $p^3$ .

P^3 is divisible by 1, P, P^2, P^3

Hence there are 4 factors

Prime factorisation of $p^3$ would still be $p^3$ as p is prime. The number of factors of an integer we assume $a^n$ , is $n+1$ . This also works on $(a^n)*(b^m)$ , number of factors would still be $(n+1)(m+1)$ .

If p^3 is prime, It has 4 factors namely 1, p, p^2, p^3

The factors of p^3 are: p, p^2, p^3, and 1. Hence 4 distinct factors.