Question

Cho x=1 + \(\sqrt[3]{5}\) + \(\sqrt[3]{25}\). Tính P=(x³-3x²-12x-15)¹⁰+2018.

Answer 1

\(x=1+\sqrt[3]{5}+\sqrt[3]{25}\Rightarrow x-1=\sqrt[3]{5}+\sqrt[3]{25}\)

\(\Rightarrow\left(x-1\right)^3=5+25+3.\sqrt[3]{5.25}\left(\sqrt[3]{5}+\sqrt[3]{25}\right)=30+15\left(\sqrt[3]{5}+\sqrt[3]{25}\right)\)

\(\Rightarrow\left(x-1\right)^3=30+15\left(x-1\right)=15+15x\)

Ta có:

\(P=\left(x^3-3x^2+3x-1-15x-14\right)^{10}+2018\)

\(P=\left(\left(x-1\right)^3-15x-14\right)^{10}+2018=\left(15+15x-15x-14\right)^{10}+2018\)

\(\Rightarrow P=1^{10}+2018=1+2018=2019\)