Đặt \(\frac{a}{b}=\frac{b}{3c}=\frac{c}{9a}=k\)
Ta có: \(\frac{a}{b}=\frac{b}{3c}=\frac{c}{9a}\)
\(\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{3c}\right)^3=\left(\frac{c}{9a}\right)^3=\frac{a.b.c}{b.3c.9a}=\frac{1}{27}=k^3\)
\(\Rightarrow k=\frac{1}{3}\)
Ta có: \(\frac{b}{3c}=\frac{1}{3}\)
\(\Rightarrow b=\frac{1}{3}.3c=c\)
Vậy \(b=c\left(đpcm\right)\)