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...

Question

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.

Answer 1

$\left(1+x^{2}+x^{4}\right)^{8}={^{16} C_{0}}+{^{16}C_{1}} x^{2}+{^{16} C_{2}} x^{4}+\cdots +{^{16} C_{16}} x^{32}$

I took the equation given by @ashish,also it is wrong as pointed out by@Tavish

put $x=\omega$

$(1+\omega^2+\omega^4)^8={^{16} C_{0}}+{^{16} C_{1}} \omega^2+{^{16} C_{2}} \omega^4+\cdots$

$0=\omega^2({^{16} {C}_{1}}+{^{16} {C}_{4}}+{^{16} {C}_{7}}+{^{16} {C}_{10}}+{^{16} {C}_{13}}+{^{16}{C}_{16}})+\omega(^{16} {C}_{2}+\cdots)+(^{16}{C}_{0}+\cdots)$ ($1$)

by comparing with equation $\omega^2+\omega+1=0$

the coeffecients of $\omega,\omega^2,1$ are equal(this can also be proved by taking imaginary and real parts in above equation).so as You said you got $(^{16}{C}_{0}+\cdots)$the same value is for the required series.

all the numbers are as per the question so the values may not be equal but this answer is to know the method

Answer 2

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 $$ a = b= c= 3^7$$