Show that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$

Let $$\theta=2\pi/7$$, show that $$1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$$

I have found that $$\cos(4\theta) = \cos(8\pi/7) = \cos(6\pi/7) = \cos(\theta)$$ and, similarly, $$\cos(5\theta) = \cos(2\theta)$$ and $$\cos(6\theta) = \cos(\theta)$$.Thus the expression $$1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\cos(4\theta)+\cos(5\theta)+\cos(6\theta) = 0$$ can be written as

$$1+2\cos(\theta)+2\cos(2\theta)+2\cos(3\theta)= 0$$

Where do I get that $$\cos(\theta)$$ satisfies the polynomial $$8x^3+4x^2-4x-1$$.

But at the moment I don't see how it can help me with what I need. Is it the correct way? If not, could you give me a hint of where to go?

If you can use the property that $$\cos(x) = Re(e^{i x})$$, where $$Re(z)$$ is the real part of $$z$$, then:
\begin{align*} \sum_{k=0}^6 \cos(k\theta) &= Re\left(\sum_{k=0}^6 e^{ik\theta}\right) \\ &= Re\left(\frac{1-e^{i7\theta}}{1-e^{i\theta}}\right) \\ &= Re\left(\frac{1-e^{i2\pi}}{1-e^{i\frac{2\pi}{7}}}\right) \\ &= 0 \end{align*} as $$e^{i2\pi} = 1$$.