已知\(m=\sqrt[3]{4}+\sqrt[3]{2}+1\),求\(\frac{1}{m^3}+\frac{3}{m^2}+\frac{3}{m}\)的值。
用到的知识点有：
1、\(a^3-1=(a-1)(a^2+b+1)\)
2、\((a+1)^3 =a^3+3a^2+3a+1^3\)
\[
\begin{align*}
已知：m=\sqrt[3]{4}+\sqrt[3]{2}+1 \\
& m=(\sqrt[3]{2})^2+\sqrt[3]{2}+1 \\
& 设 t = \sqrt[3]{2} \\
& m =t^2 + t +1 \\
& 两边同乘以 t-1 \\
& (t-1)\cdot m = (t-1)\cdot(t^2+t+1) \\
& (t-1)\cdot m = t^3 -1 \\
& (t-1)\cdot m = (\sqrt[3]{2})^3 -1=2-1=1 \\
& m = \frac{1}{\sqrt[3]{2}-1} \\
得：\frac{1}{m} = \sqrt[3]{2}-1 \\
\\
\frac{1}{m^3}+\frac{3}{m^2}+\frac{3}{m} \\
& = (\frac{1}{m})^3 +3(\frac{1}{m})^2+3(\frac{1}{m}) \\
& = (\frac{1}{m})^3 +3(\frac{1}{m})^2+3(\frac{1}{m})+1-1 \\
& = (\frac{1}{m}+1)^3 -1 \\
& = (\sqrt[3]{2}-1+1)^3 -1 \\
& = (\sqrt[3]{2})^3-1 \\
& = 2-1=1 \\
\end{align*}
\]