Regarding the structure of certain minimal polynomials My question is: is the following statement true? If so, how might I go about proving it? 

Let $m$ be an integer not less than 1, let $F = GF(2^{6m})$, and let $L = GF(2^{2m})$. Let $\gamma$ be a primitive element of $L$, and let $\beta \in F$ be such that $\beta^3 = \gamma$. If $\phi_a(x) = c_0x^{6m}+c_1x^{6m-1}+...+c_{6m-1}x+c_{6m}$ is the minimal polynomial of an element $a \in \beta L$, then 
$$c_i =  0 \hspace{5mm} \text{if $i \not\equiv 0 \pmod{3}$}$$
 A: The following sketch is the best I can muster right away. It probably isn't the cleanest route to the destination, but it leads to a proof.
Assume that $z=\gamma^j$ is an arbitrary non-zero element of $L$. Then


*

*$\beta z=\beta^{1+3j}$, and

*$(\beta z)^3=\gamma^{1+3j}\in L$.


Let $\phi(x)$ be the minimal polynomial of $\gamma^{1+3j}\in L$. Then


*

*$a=\beta z$ is a zero of the polynomial $\phi(x^3)$, and

*the coefficients of terms of $\phi(x^3)$ of degree $\not\equiv0\pmod3$ are all zero.


This proves your observation, if we can show that $\phi(x^3)$ has the correct degree ($=\deg\phi_a(x)$). Because $\deg\phi(x)$ is always a factor of $2m$, we are done, if we have that extra bit of information that $\deg\phi_a(x)=6m$.
Even if that is not the case we can make the following deductions (leaving them sketchy as per your request):


*

*We have $\deg\phi(x)$ is equal to the degree of the extension $GF(2)[a^3]$ over $GF(2)$. 

*Similarly $\deg\phi_a(x)$ is equal to the degree of the extension $GF(2)[a]$ over $GF(2)$.

*Here $GF(2)[a^3]$ is a subfield of $L$.

*But $GF(2)[a]$ is not a subfield of $L$. It is a subfield of $F$ though.

*Clearly $GF(2)[a^3]$ is a subfield of $GF(2)[a]$.

*Putting these bits together shows that $GF(2)[a]$ is a cubic extension of $GF(2)[a^3]$. 

*Therefore $\deg\phi_a(x)=3\deg\phi(x)$, and the claim follows as above.


The next to last bullet is a bit subtler than the rest. Remember that the cardinality of a subfield of $F$ determines it uniquely.
