Câu 1:
Đặt \(a=\sqrt[3]{4+\sqrt{15}},\) \(b=\sqrt[3]{4-\sqrt{15}}.\) Ta có:
\(a^3-b^3=4+\sqrt{15}-4+\sqrt{15}=2\sqrt{15}\)
\(ab=\sqrt[3]{4+\sqrt{15}}.\sqrt[3]{4-\sqrt{15}}=\sqrt[3]{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=\sqrt[3]{16-15}=1\)
Ta có:
\(P=x^3+3x=\left(a-b\right)^3+3\left(a-b\right)=a^3-b^3-3ab\left(a-b\right)+3\left(a-b\right)\)
\(=2\sqrt{15}-3\left(a-b\right)+3\left(a-b\right)\) (vì \(a^3-b^3=2\sqrt{15},\) \(ab=1\))
\(=2\sqrt{15}\)
Vậy \(P=2\sqrt{15}\)
Câu 2:
Ta có: \(x=\sqrt[3]{2}-2\) \(\Leftrightarrow\) \(x+2=\sqrt[3]{2}\) \(\Rightarrow\) \(\left(x+2\right)^3=\left(\sqrt[3]{2}\right)^3\) \(\Leftrightarrow\) \(x^3+6x^2+12x+8=2\) \(\Leftrightarrow\) \(x^3+6x^2+12x+6=0\)
Vậy \(P=x^3+6x^2+12x+1=\left(x^3+6x^2+12x+6\right)-5=-5\) (vì \(x^3+6x^2+12x+6=0\))
Câu 3:
Ta có: \(x=2\sqrt[3]{2}-2\) \(\Leftrightarrow\) \(x+2=2\sqrt[3]{2}\) \(\Rightarrow\) \(\left(x+2\right)^3=\left(2\sqrt[3]{2}\right)^3\) \(\Leftrightarrow\) \(x^3+6x^2+12x+8=16\) \(\Leftrightarrow\) \(x^3+6x^2+12x-8=0\)
Vậy \(P=x^4+8x^3+24x^2+16x+2018\)
\(=\left(x^4+6x^3+12x^2-8x\right)+\left(2x^3+12x^2+24x-16\right)+2034\)
\(=x\left(x^3+6x^2+12x-8\right)+2\left(x^3+6x^2+12x-8\right)+2034\)
\(=2034\) (vì \(x^3+6x^2+12x-8=0\))
