Question

Cho \(b=\sqrt[3]{2020}\). Tính \(Q=\sqrt[3]{\frac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}}{2}}+\sqrt[3]{\frac{b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}}\)

Answer 1

\(Q=\sqrt[3]{\frac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}}{2}}+\sqrt[3]{\frac{b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}}\)

\(\Leftrightarrow Q^3=b^3-3b+3Q\sqrt[3]{\frac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}}{2}}.\sqrt[3]{\frac{b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}}\)

\(\Leftrightarrow Q^3=b^3-3b+3Q\)

\(\Leftrightarrow\left(Q-b\right)\left(Q^2+Qb+b^2-3\right)=0\)

Dễ thấy \(Q^2+Qb+b^2-3>0\)

\(\Rightarrow Q=b=\sqrt[3]{2020}\)