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If $x=y \cos \left(\frac{2 \pi}{3}\right)=z \cos \left(\frac{4 \pi}{3}\right)$, then what is
$x y+y z+z x$ equal to?
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$x y+y z+z x$ equal to?
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Let $x=y \cos \left(\frac{2 \pi}{3}\right)=z \cos \left(\frac{4 \pi}{3}\right)$
$\Rightarrow x=y \cos \left(\pi-\frac{\pi}{3}\right)=-y \cos \frac{\pi}{3}=\frac{-y}{2}$
and $x=z \cos \left(\pi+\frac{\pi}{3}\right)=-z \cos \frac{\pi}{3}=\frac{-z}{2}$
from (1) and (2)
$\frac{-y}{2}=\frac{-z}{2} \Rightarrow y=z$
$\begin{aligned} \text { Thus, } x y+y z+z x &=z x+z^{2}+x z=2 x z+z^{2} \\ &=-y \cdot(y)+y^{2}=-y^{2}+y^{2}=0
$\Rightarrow x=y \cos \left(\pi-\frac{\pi}{3}\right)=-y \cos \frac{\pi}{3}=\frac{-y}{2}$
and $x=z \cos \left(\pi+\frac{\pi}{3}\right)=-z \cos \frac{\pi}{3}=\frac{-z}{2}$
from (1) and (2)
$\frac{-y}{2}=\frac{-z}{2} \Rightarrow y=z$
$\begin{aligned} \text { Thus, } x y+y z+z x &=z x+z^{2}+x z=2 x z+z^{2} \\ &=-y \cdot(y)+y^{2}=-y^{2}+y^{2}=0
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