For each $k\geq 0$, describe $H^k(X,\mathcal{O}_X)$ where $X:=\text{Spec}(k[x,y, z])\setminus\{(x,y,z)\}$.
The first definition of cohomology I've learned involves injective resolutions, which I have no idea how to apply here.
I've read some authors who claimed that Cech cohomology is often useful to compute sheaf cohomology in real life, so I decided to take that road.
If $A:=k[x,y,z]$ and $Y:=\text{Spec}(A)$, I've thought about using the affine open cover $U_x:=Y\setminus V(x)$, $U_y:=Y\setminus V(y)$ and $U_z:=Y\setminus V(z)$.
That way, $\mathcal{O}_X(U_x)=A_x$, $\mathcal{O}_X(U_y)=A_y$ and $\mathcal{O}_X(U_z)=A_z$ and consequently: $$C^0:=C^0(\mathcal{O}_X)=A_x\times A_y\times A_z$$ $$C^1=A_{xy}\times A_{xz}\times A_{yz}$$ $$C^2=A_{xyz}$$
I've checked that $\left(\frac{a}{x^n},\frac{b}{y^m},\frac{c}{z^\ell}\right)\in\ker(d^0)\Leftrightarrow\frac{a}{x^n}=\frac{b}{x^m}=\frac{c}{z^\ell}\in k[x,y,z]$, so $\ker(d^0)\simeq A$ and $H^0(X,\mathcal{O}_X)=A$.
Now, determining $H^1(X,\mathcal{O}_X)=\ker(d^1)/\text{im}(d^0)$ and $H^2(X,\mathcal{O}_X)=C^1/\text{im}(d^1)$ is considerably more complicated and made me wonder whether or not this is the best option.
I've also tried the simpler case $\text{Spec}(k[x,y])\setminus\{(x,y)\}$, and even then I found hard to describe $H^1(X,\mathcal{O}_X)$ explicitly.
This looks like a standard problem, so I can't help but wonder if there isn't a simpler approach which could work even for the general case $\text{Spec}(k[x_1,...,x_n])\setminus\{(x_1,...,x_n)\}$.
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