Question

Cho \(x=\dfrac{1}{3}\left(1+\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)\). Tính giá trị của biểu thức: \(M=\left(9x^3-9x^2-3\right)^2\)

Answer 1

Do \(12=\sqrt{144}>\sqrt{135}\) nên \(x>0\)

Đặt \(a=\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\) \(\Rightarrow x=\dfrac{1}{3}\left(a+1\right)\)

\(a^3=8+3\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)=8+3a\)

Ta có: \(x=\dfrac{1}{3}\left(a+1\right)\Rightarrow3x=a+1\Rightarrow9x^2=a^2+2a+1\)

Lại có: \(x^3=\dfrac{1}{27}\left(a+1\right)^3\Leftrightarrow9x^3=\dfrac{1}{3}\left(a^3+3a^2+3a+1\right)\)

\(\Leftrightarrow9x^3=\dfrac{1}{3}\left(8+3a+3a^2+3a+1\right)=a^2+2a+3\)

\(\Rightarrow M=\left(a^2+2a+3-a^2-2a-1-3\right)^2=\left(-1\right)^2=1\)