A polynomial $f(x)$ of degree larger than $5$ may not be solvable. But if we take, for example, $$ f(x)=x^8+x^6+x^4+2, $$ we can see by introducing a new variable $u=x^2$ (by observation) and it becomes a quartic equation and we can solve it by radicals. My question is whether one can know this change of variable should be done by just studying its Galois group?
In general, if a polynomial $f(x)$ is solvable, I would like to ask does Galois theory provide us methods to solve $f(x)$(maybe some algorithms) when we have a well understanding of the Galois group of $f$? For example, rediscover how one can solve cubic and quartic equations.
For your case, $f(x)=0$, there are additional equations which the 8 roots obey. One might write $$x_1+x_2=0,\,x_3+x_4=0\\ x_5+x_6=0,\,x_7+x_8=0 $$The members of the Galois group must preserve these equations. The Galois group is approximately $P_4 \times Z_2 \times Z_2 \times Z_2 \times Z_2 $. The $Z_2$'s come from swapping $x_1$ with $x_2$, etc. and the $P_4$ comes from swapping entire pairs of roots. The $Z_2$'s make up normal subgroups, the quotient group is $P_4$. The route to the roots begins by finding polynomials invariant under the subgroup, but not the quotient group. The $x_i^2$'s are a collection that is invariant under the $z_2$ subgroups. The quartic's, which I describe next are more interesting.
We seek the 4 roots $\{z_1, z_2, z_3, z_4\}$ of $\, 0=a_0+a_1 z+a_2 z^2+a_3 z^3+a_4 z^4$. The Galois group is $P_4$ and the largest normal subgroup is $A_4$ the set of even permutations to "solve" a polynomial one looks for new polynomials that are invariant under the subgroup, but which transform under the quotient group. Define $$R=(x_1-x_2)(x1-x_3)(x_1-x_4)(x_2-x_3)(x_2-x_4)(x_3-x_4)$$. R changes sign under odd permutations, and is invariant under $A_4$. $R^2$ is invariant under $P_4$ and can be expressed as a multinomial in the $\{a_i\}$ The largest normal subgroup of $A_4$ is the group of permutations that involve two disjoint interchanges, such as $x_1 \leftrightarrow x_2$ and $x_3 \leftrightarrow x_4$. This group has 4 elements including the identity and is a simple product $Z_2 \times Z_2$. Call this group G. The eigenvectors of this subgroup can be written $y_1 =x_1+x_2-x_3-x_4$, $y_2 = x_1+x_3-x_2-x_4$ and $y_3 =x_1+x_4-x_2-x_3$. These eigenvectors are the ones that are invariant under one of the $Z_2$'s but change sign under the others. Note that G can change the sign of any 2 y's, but not 1 or 3. This suggests that $y_1 y_2 y_3$ is some kind of invariant and it can be shown with some difficulty that $y_1 y_2 y_3= -8 a_1+4 a_2 a_3 - a_3^3$. The Galois group is an excellent guide to finding good intermediate variables, but the algebra can be complicated. The next step is to find variables invariant under G but not $A_4$. Good choices are $$u_1=(x_1-x_2) (x_3-x_4)\\ u_2=(x_3-x_1) (x_2-x_4) \\ u_3=(x_1-x_4) (x_3-x_2)$$. $A_4$ rearranges the $\{ u_i \}$ but leaves the set unchanged. The $u_i$'s obey the following:$$u_1+u_2+u_3=0\\ u_1 u_2+u_1 u_3+u_2 u_3 = -12 a_0-a_2^2+3a_1a_3\\ u_1 u_2 u_3 = R\\ (u_1-u_2)(u_1-u_3)(u_2-u_3)=27 a1^2 - 72 a_0 a_2 + 2 a_2^3 - 9 a_1 a_2 a_3 + 27 a_0 a_3^2$$The last equation limits the allowed order of roots. These $u_i$'s can be found by solving a cubic.
Consider the y's, and note $$y_1^2=(-8 a_2 + 3 a_3^2 - 4 u_1 - 8 u_2)/3\\ y_2^2 = (-8 a_2 + 3 a_3^2 + 8 u_1 + 4 u_2)/3\\ y_3^2 = (-8 a_2 + 3 a_3^2 - 4 u_1 + 4 u_2)/3 $$ remember, $ y_1$ $ y_2 $ $ y_3 $ is known, so we cannot arbitrarily choose the branch of the square roots.