Đặt \(a=\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\) \(\Rightarrow x=\dfrac{1}{3}\left(a+1\right)\)
\(\Rightarrow3x=a+1\Rightarrow9x^2=a^2+2a+1\) (1)
\(x^3=\dfrac{1}{27}\left(a+1\right)^3=\dfrac{1}{27}\left(a^3+3a^2+3a+1\right)\)
Ta có:
\(a^3=\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)^3\)
\(\Rightarrow a^3=\dfrac{24}{3}+3\sqrt[3]{\dfrac{\left(12+\sqrt{135}\right)\left(12-\sqrt{135}\right)}{9}}.\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)\)
\(\Rightarrow a^3=8+3a\)
\(\Rightarrow x^3=\dfrac{1}{27}\left(8+3a+3a^2+3a+1\right)=\dfrac{1}{9}\left(a^2+2a+3\right)\)
\(\Rightarrow9x^3=a^2+2a+3\) (2)
Thay (1), (2) vào M ta được:
\(M=\left(9x^3-9x^2-3\right)^2=\left(a^2+2a+3-\left(a^2+2a+1\right)-3\right)^2\)
\(\Rightarrow M=\left(-1\right)^2=1\)
