Question:
If $a=b \cos \frac{2 \pi}{3}=c \cos \frac{4 \pi}{3}$, then write the value of $a b+b c+c a$.
Solution:
$a=b \cos 120^{\circ}=c \cos 240^{\circ}$
$\Rightarrow a=-\frac{1}{2} b=-\frac{1}{2} c$
Therefore,
$a b+b c+c a=\frac{-1}{2} b \times b+b \times b+b \times \frac{-1}{2} b$
$=-b^{2}+b^{2}$
$=0$
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