Question

Cho \(x=1+\sqrt[3]{2}+\sqrt[3]{4}\)

Tính \(A=x^5-4x^4+x^3-x^2-2x+2019\)

Answer 1

\(x-1=\sqrt[3]{2}+\sqrt[3]{4}\)

\(\Rightarrow\left(x-1\right)^3=6+3\sqrt[3]{8}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)

\(\Rightarrow x^3-3x^2+3x-1=6+6\left(x-1\right)\)

\(\Rightarrow x^3-3x^2-3x-1=0\)

\(A=x^2\left(x^3-3x^2-3x-1\right)-x^4+4x^3-2x+2019\)

\(=-x\left(x^3-3x^2-3x-1\right)+x^3-3x^2-3x+2019\)

\(=1+2019=2020\)