Prime factors

If $P^3+2P^2+P=P(P+1)^2$ has $42$ divisors, then $(P+1)^2$ has $21$ divisors. Let the powers of the prime factors of $P+1$ be $a,b,c,\dots$ . Then we can write;

$(P+1)^2=(P_1^{a}\cdot P_2^{b}\cdot P_3^{c}\dots)^2$

Therefore, $(2a+1)(2b+1)(2c+1)\dots=21$

This is possible if $P+1$ has $2$ factors, $2a+1=3$ and $2b+1=7$ .

Which means that, $a=1, b=3$

Hence $P+1$ is of the form $(P_1\cdot P_2^3)$

Since $24=2^3\cdot3$ is the smallest possible case, we get $P=23$