I am trying to understand how to compute, explicitly, the top self intersection of a divisor inside a variety which is not a surface. I am excluding the surface case because there I can use the intersection product among curves and divisors.
Let's consider a specific example: let $X$ be the blow-up of $\mathbb{P}^3$ along a line $l$. Suppose I want to compute $(-K_X)^3$, where $-K_X$ is the canonical divisor. I know that $-K_X=4H-E$, where $H$ is the pullback of the hyperplane class in $\mathbb{P}^3$ and $E$ is the exceptional divisor associated to the blow-up. I guess I can expand it as
$$(-K_X)^3=(4H-E)^3=64H^3-E^3-48H^2E+12HE^2,$$
but then I'm lost, because I don't know the value of any of these pieces. Arguing as in the surface case, I guess that $H^3=1$, $E^3=-1$, but $H^2E,EH^2$ don't even seem to be well-defined (and I'm just guessing, there is no rigorous thinking behind it).
Any help would be much appreciated, I'm feeling really lost.
One can use the projection formula. For instance, if $p$ is the projection to $\mathbb{P}^3$, one has $$ (p^*H)^3 = p^*(H^3) \cdot [X] = H^3 \cdot p_*[X] = H^3 \cdot [\mathbb{P}^3] = 1. $$ Similarly, if $i : E \to X$ is the embedding, then $$ H^2 \cdot E = H^2 \cdot i_*[E] = i^*(H^2) \cdot [E] = (i^*H)^2 = 0 $$ since $i^*H = 0$, and analogously $$ H \cdot E^2 = H \cdot E \cdot i_*[E] = i^*(H \cdot E) \cdot [E] = (i^*H) \cdot (i^*E) = 0. $$ Finally, $$ E^3 = E^2 \cdot i_*[E] = i^*(E^2) \cdot [E] = (i^*E)^2 = (-h)^2 = 1, $$ since $i^*E = -h$, where $h$ is the hyperplane class of $E \cong \mathbb{P}^2$.