\(a=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\)
\(\Leftrightarrow a^3=\left(\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\right)^3\)
\(\Leftrightarrow a^3=\left(2-\sqrt{3}\right)+3\sqrt[3]{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}.\left(\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\right)+\left(2+\sqrt{3}\right)\)
\(\Leftrightarrow a^3=4+3\sqrt[3]{4-3}.a\)
\(\Leftrightarrow a^3=4+3a\)
\(\Leftrightarrow a^3-3a=4\)
\(\Leftrightarrow a\left(a^2-2\right)=4\)
\(\Leftrightarrow a^2-2=\dfrac{4}{a}\left(a\ne0\right)\)
\(M=\dfrac{64}{\left(a^2-3\right)^3}-3a\)
\(=\dfrac{64}{\left(\dfrac{4}{a}\right)^3}-3a\)
\(=\dfrac{64}{\dfrac{64}{a^3}}-3a=a^3-3a=4\)
Vì 4 là số chính phương nên M cũng là số chính phương.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2+\sqrt[]{3}}=x\\\sqrt[3]{2-\sqrt[]{3}}=y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^3+y^3=4\\xy=\sqrt[3]{\left(2+\sqrt[]{3}\right)\left(2-\sqrt[]{3}\right)}=1\end{matrix}\right.\)
\(a^2=x^2+y^2+2xy\)
\(\Rightarrow M=\left(\dfrac{4}{a^2-3}\right)^3-3a=\left(\dfrac{x^3+y^3}{x^2+y^2+2xy-3xy}\right)^3-3\left(x+y\right)\)
\(=\left[\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x^2-xy+y^2}\right]^3-3\left(x+y\right)\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)=x^3+y^3=4\) là SCP
![\displaystyle a=\sqrt[3]{{2-\sqrt{3}}}+\sqrt[3]{{2+\sqrt{3}}}](https://toancap2.net/wp-content/ql-cache/quicklatex.com-f97fff04475ef931060f897e441e2052_l3.svg)
. Chứng minh 
là số nguyên
