\(M=\left(9x^3-9x^2-3\right)^2\)
Hình như tính cái này
Đặt \(a=\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)
\(\Rightarrow a^3=\left(\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\right)^3\)
Có (a+b)^3=a^3+b^3+3ab(a+b)
\(\Rightarrow a^3=4+\sqrt{15}+4-\sqrt{15}+3\sqrt[3]{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}a\)
\(\Rightarrow a^3=8+3a\Rightarrow a^3-3a-8=0\)-> khó
\(3x=1+\sqrt[3]{\frac{12+3\sqrt[]{15}}{3}}+\sqrt[3]{\frac{12-3\sqrt[]{15}}{3}}\)
\(\Leftrightarrow3x-1=\sqrt[3]{4+\sqrt[]{15}}+\sqrt[3]{4-\sqrt[]{15}}\)
\(\Leftrightarrow\left(3x-1\right)^3=4+\sqrt[]{15}+4-\sqrt[]{45}+3\sqrt[3]{\left(4-\sqrt[]{15}\right)\left(4+\sqrt[]{15}\right)}\left(3x-1\right)\)
<=> \(\left(3x-1\right)^3=8+3\left(3x-1\right)\Leftrightarrow\left(3x-1\right)^3-3\left(3x-1\right)-8=0\)
Khai chuyển rút gọn ta đc
\(27x^3-27x^2-6=0\Leftrightarrow9x^3-9x^2-2=0\Leftrightarrow9x^3-9x^2-3=-1\)
<=> M = ( - 1 )^2 = 1