If $P=99$, then $P(P^2+3P+3)=?$

Let's expand the given algebraic expression:
$$E=P(P^2+3P+3)=P^3+3P^2+3P$$

We know from standard algebraic expansions that:

$$(P+1{)}^3=P^3+3P^2+3P+1$$

Subtracting $1$ from both sides gives:

$$P^3+3P^2+3P=(P+1{)}^3-1$$

Now, substitute the value $P=99$:

$$E=(99+1{)}^3-1$$

$$E=(100{)}^3-1=1,000,000-1=999,999$$