If $\left(1+x^{2}+x^{4}\right)^{8}={^{16} C_{0}}+{^{16}C_{1}} x^{2}+{^{16} C_{2}} x^{4}+\cdots +{^{16} C_{16}} x^{32}$, how to prove... If I am given that $$\left(1+x^{2}+x^{4}\right)^{8}={^{16} C_{0}}+{^{16}C_{1}} x^{2}+{^{16} C_{2}} x^{4}+\cdots +{^{16} C_{16}} x^{32},$$
how can I prove that $${^{16} {C}_{1}}+{^{16} {C}_{4}}+{^{16} {C}_{7}}+{^{16} {C}_{10}}+{^{16} {C}_{13}}+{^{16}{C}_{16}}=3^{7}$$ is true?
My approach: I usually try to evaluate these by putting $x =\sqrt{-1}, \sqrt[3]{-1}$ however I'm not sure how to approach this. Please suggest any hints or methods as to how to solve this particular and in general how to approach these types of expressions.

The difference between the powers of $x^6$ between subsequent coefficients, so I suggest dividing this polynomial by $x^6-1$ and checking the remainder. If $$ (1+x^2+x^4)^8 = {}^{16}C_0 +{}^{16}C_1 x^2 + \dots + {}^{16}C_{16} x^{32} = w(x) (x^6-1) + r(x)$$
we have
$$r(x) = a + b x^2 + c x^4$$
where
$$ a = {}^{16}C_0 +{}^{16}C_3 + \dots + {}^{16}C_{30} $$
$$ b = {}^{16}C_1 +{}^{16}C_4+ \dots + {}^{16}C_{31} $$
$$ c = {}^{16}C_2 +{}^{16}C_5 + \dots + {}^{16}C_{32} $$
If you put $x^2 \in\{1, (-1)^\frac23, (-1)^\frac43\}$ you have $x^6-1 =0 $, so
$$a + b+ c = (1+1+1)^8 = 3^8$$
$$a + b (-1)^\frac23 + c (-1)^\frac43 = (1+(-1)^\frac23+(-1)^\frac43)^8 = 0$$
$$a + b (-1)^\frac43 + c (-1)^\frac83 = (1+(-1)^\frac43+(-1)^\frac83)^8 = 0$$
Solving these equations gives you
$$ a = b= c= 3^7$$
