can we have split field of the real number and how i get the extension dimension? what the spilt field of x^3-8? I think I cannot split the real number,is that correct? what about x^3-2 ,is the extension dimension will be either 6 or 3? In the beginning I think it will be 3 extension dimension ,since we have three roots. Then I look for the answer it is 6 extension dimension why?

The polynomial $x^3-8$ has the rational root $2$. Divide the polynomial by $x-2$. We get $x^2+2x+4$, an irreducible quadratic. So the splitting field of $x^3-8$ has degree $2$ over the rationals.

For the polynomial $x^3-2$, adjoin $\sqrt[3]{2}$ to the rationals. Since $x^3-2$ is irreducible of degree $3$, the field $E=\mathbb{Q}(\sqrt[3]{2})$ has degree $3$ over the rationals. The splitting field of $x^3-2$ is not $E$, since $x^3-2$ has some complex roots. Imagine dividing $x^3-2$ by $x-\sqrt[3]{2}$. We get an irreducible polynomial of degree $2$, so its splitting field $F$ has degree $2$ over $E$. It follows that the degree of $F$ over $\mathbb{Q}$ is $(3)(2)$.