\(\frac{a}{b}=\frac{b}{3c}=\frac{c}{9a}\)
\(\Leftrightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{3c}\right)^3=\left(\frac{c}{9a}\right)^3=\left(\frac{a.b.c}{b.3c.c.9a}\right)=\frac{1}{27}=k^3\)
\(\Leftrightarrow k=\left(\frac{1}{27}\div\frac{1}{27}\right)\div3=\frac{1}{3}\)
\(\Leftrightarrow\frac{b}{3c}=\frac{1}{3}\)
Vậy \(\Rightarrow b=c\left(đpcm\right)\)