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If $\cosh x=\frac{4}{3}$, then $3 \cosh x+3^2 \cosh 2 x+3^3 \cosh 3 x=$
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$175$
$\cos h x=\frac{4}{3}$
$\begin{aligned} & \cos h^2 x-\sin h^2 x=1 \\ & \begin{aligned} & \Rightarrow \sin h^2 x=\left(\frac{4}{3}\right)^2-1=\frac{7}{9} \Rightarrow \sin h x=\frac{\sqrt{7}}{3} \\ & \cos h 2 x=\cos h^2 x+\sin h^2 x=\frac{16}{9}+\frac{7}{9}=\frac{23}{9} \\ & \cos h 3 x=4 \cos h^3 x-3 \cos h x=4\left(\frac{4}{3}\right)^3-3\left(\frac{4}{3}\right) \\ & \therefore \quad=\frac{256}{27}-\frac{12}{3}=\frac{148}{27} \\ &=3\left(\frac{4}{3}\right)+9\left(\frac{23}{9}\right)+27\left(\frac{148}{27}\right)=4+23+148=175 .\end{aligned}\end{aligned}$
$\begin{aligned} & \cos h^2 x-\sin h^2 x=1 \\ & \begin{aligned} & \Rightarrow \sin h^2 x=\left(\frac{4}{3}\right)^2-1=\frac{7}{9} \Rightarrow \sin h x=\frac{\sqrt{7}}{3} \\ & \cos h 2 x=\cos h^2 x+\sin h^2 x=\frac{16}{9}+\frac{7}{9}=\frac{23}{9} \\ & \cos h 3 x=4 \cos h^3 x-3 \cos h x=4\left(\frac{4}{3}\right)^3-3\left(\frac{4}{3}\right) \\ & \therefore \quad=\frac{256}{27}-\frac{12}{3}=\frac{148}{27} \\ &=3\left(\frac{4}{3}\right)+9\left(\frac{23}{9}\right)+27\left(\frac{148}{27}\right)=4+23+148=175 .\end{aligned}\end{aligned}$
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