If \(3 \mathrm{f}(x)-\mathrm{f}\left(\frac{1}{x}\right)=8 \log _2 x^3, x>0\), then \(\mathrm{f}(2), \mathrm{f}(4)\) \(\mathrm{f}(8)\) are in

\(
\begin{aligned}
& 3 f(x)-f\left(\frac{1}{x}\right)=8 \log _2 x^3 \\
& \Rightarrow 3 \mathrm{f}\left(\frac{1}{x}\right)-\mathrm{f}(x)=8 \log _2\left(\frac{1}{x}\right)^3 .
\end{aligned}
\)
From (i) and (ii), we get
\(
\begin{aligned}
& 8 \mathrm{f}(x)=24 \log _2 x^3+8 \log _2\left(\frac{1}{x}\right)^3 \\
& \Rightarrow 8 \mathrm{f}(x)=72 \log _2 x-24 \log _2 x \\
& \Rightarrow 8 \mathrm{f}(x)=48 \log _2 x \\
& \Rightarrow \mathrm{f}(x)=6 \log _2 x
\end{aligned}
\)
\(\therefore f^{\prime}(2)=6 \log _2 2=6 \)
\( f^{\prime}(4)=6 \log _2 4=12 \)
\( f^{(}(8)=6 \log _2 8=18 \)
\( \therefore f(2), f(4), f(8) \text { are in A.P. }\)