Lời giải:
Ta có: \(f(x)=x^6+2x^3+1=(x^3+1)^2\)
\(\Rightarrow \left\{\begin{matrix} f(\sqrt[3]{3+2\sqrt{2}})=(3+2\sqrt{2}+1)^2=(4+2\sqrt{2})^2\\ f(\sqrt{2})=(2\sqrt{2}+1)^2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} f(\sqrt[3]{3+2\sqrt{2}})=(4+2\sqrt{2})^2\\ 4f(\sqrt{2})=(4\sqrt{2}+2)^2\end{matrix}\right.\)
\(\Rightarrow f(\sqrt[3]{3+2\sqrt{2}})-4f(\sqrt{2})=(4+2\sqrt{2}-4\sqrt{2}-2)(4+2\sqrt{2}+4\sqrt{2}+2)\)
\(=(2-2\sqrt{2})(6+6\sqrt{2})=12(1-\sqrt{2})(1+\sqrt{2})=-12\)