If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$ THE PROBLEM: If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$
MY THOUGHT PROCESS: We have to prove that $ab+bc+ca=0$. 
One method using which we can do this is, if we can somehow obtain the equation $k(ab+bc+ca)=0$ we can deduce that $ab+bc+ca=0$ using the zero product rule. 
Another method to do this would be to obtain an expression which has $(ab+bc+ca)$ in it. The identity $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ has $2(ab+bc+ca)$ in it. Therefore if we can somehow show that $(a+b+c)^2=a^2+b^2+c^2$ our task would be over.
MY ATTEMPT: I have proved the result using the first approach. I tried to do it using the second one but could not proceed far. I was facing problems showing that $(a+b+c)^2=a^2+b^2+c^2$. If i get any help i shall be very grateful.

I don't know whether I am just repeating your first solution.
Let $a\cos\theta=k$. Then
\begin{align*}
k(ab+bc+ca)&=abc\cos\left(\theta+\frac{4\pi}{3}\right)+abc\cos\theta+abc\cos\left(\theta+\frac{2\pi}{3}\right)\\
&=abc\left[\cos\left(\theta+\frac{2\pi}{3}\right)+2\cos\left(\theta+\frac{2\pi}{3}\right)\cos\frac{2\pi}{3}\right]\\&=0
\end{align*}
