If  $\int \frac{(3x^{10}+2x^8-2)\sqrt[4]{x^{10}+x^8+1}}{x^6}dx=f\left(x\right)+C$ (where C is constant of integration, x > 0) and  $f\left(1\right)=\frac{6\sqrt[4]{3}}{5}$ then the value of  $\frac{{\left(80f\left(2\right)\right)}^{\frac{4}{5}}}{183}$ is

$$\begin{gathered} \int \frac{(3x^{10}+2x^8-2){(x^{10}+x^8+1)}^{\frac{1}{4}}}{x^6}dx \\ =\int \frac{(3x^{10}+2x^8-2)\times {\left(x^4\right)}^{\frac{1}{4}}{(x^6+x^4+x^{-4})}^{\frac{1}{4}}}{x^6}dx \\ \int (3x^5+2x^3-2x^{-5}){(x^6+x^4+x^{-4})}^{\frac{1}{4}}dx \\ {x}^6+x^4+x^{-4}=t^4 \\ (6x^5+4x^3-4x^{-5})dx=4t^{3}dt \\ (3x^5+2x^3-2x^{-5})dx=2t^{3}dt \\ =\int \left(t\right)\left(2t^3\right)dt=\frac{2t^5}{5}+C \\ =\frac{2}{5}{[x^6+x^4+x^{-4}]}^{\frac{5}{4}}+C \\ f\left(1\right)=\frac{2}{5}{(1+1+1)}^{\frac{5}{4}}=\frac{2}{5}{\left(3\right)}^{\frac{5}{4}}=\frac{6.{\left(3\right)}^{\frac{1}{4}}}{5} \\ f\left(2\right)=\frac{2}{5}{[2^6+2^4+2^{-4}]}^{\frac{5}{4}}=\frac{2}{5}\frac{{[2^{10}+2^8+1]}^{\frac{5}{4}}}{25}=\frac{1}{80}{\left[1281\right]}^{\frac{5}{4}} \end{gathered}$$