How can I show that $(1+2\cos 2\theta)^3=7+2(6\cos 2\theta+3\cos4\theta+\cos6\theta)$ using the tensor product and the Clebsch Gordan Theorem? How can I show that $$(1+2\cos2\theta)^3=7+2(6\cos2\theta+3\cos4\theta+\cos6\theta)$$
using the tensor product and the Clebsch Gordan Theorem?
The Clebsch Gordan formula for characters of $SU(2)$ is the equality
$$\chi_m  \cdot \chi_n = \chi_{m+n} + \chi_{m+n-2} + \cdots + \chi_{|m-n|}$$
where
$$\chi_m(\theta)= e^{im\theta} + e^{i(m-2)\theta} + \cdots e^{-i m \theta}$$

So you want to express $\chi_2^3$ as a combination of $\chi_n$'s. We have 
$$\chi_2^2 = \chi_4 +\chi_2+ \chi_0$$
so 
$$\chi_2^3 = \chi_2(\chi_4 +\chi_2+ \chi_0)= (\chi_6 + \chi_4 + \chi_2)+ (\chi_4 + \chi_2 + \chi_0)+\chi_2 = \chi_6 + 2 \chi_4+ 3\chi_2+ \chi_0$$

Now we  plug in the expressions for $\chi_0$, $\chi_2$, $\chi_4$, $\chi_6$ and get the result.
We have $$(1+2\cos(2\theta))^3=1+3\cdot 2\cos(2\theta)+3(2\cos(2\theta))^2+(2\cos(2\theta))^3$$ and this is (simplified) $$-1+12\, \left( \cos \left( \theta \right)  \right) ^{2}-48\, \left( 
\cos \left( \theta \right)  \right) ^{4}+64\, \left( \cos \left( 
\theta \right)  \right) ^{6}
$$
expanding the right-Hand side we get
$$-1+12\, \left( \cos \left( \theta \right)  \right) ^{2}-48\, \left( 
\cos \left( \theta \right)  \right) ^{4}+64\, \left( \cos \left( 
\theta \right)  \right) ^{6}
$$
this is the same!
$$\cos(2x)=2\cos(x)^2-1$$
$$\cos(4x)=8\cos(x)^4-8\cos(x)^2+1$$
$$\cos(6x)=32\cos(x)^6-48\cos(x)^4+18\cos(x)^2-1$$