Question

Tính B=\(x^3-3x+1077\)

với x=\(\sqrt[3]{16-\sqrt{255}}+\dfrac{1}{\sqrt[3]{16-\sqrt{255}}}\)

Answer 1

\(x^3=\left(\sqrt[3]{16-\sqrt{255}}+\dfrac{1}{\sqrt[3]{16-\sqrt{255}}}\right)^3\)

=>\(x^3=16-\sqrt{255}+\dfrac{1}{16-\sqrt{255}}+3\cdot x\cdot\sqrt[3]{\left(16-\sqrt{255}\right)\cdot\dfrac{1}{16-\sqrt{255}}}\)

=>\(x^3=16-\sqrt{255}+\dfrac{1}{16-\sqrt{255}}+3x\)

=>\(x^3-3x=16-\sqrt{255}+\dfrac{1}{16-\sqrt{255}}\)

=>\(B=16-\sqrt{255}+\dfrac{1}{16-\sqrt{255}}+1077\)

\(=1093-\sqrt{255}+\dfrac{16+\sqrt{255}}{1}\)

\(=1093-\sqrt{255}+16+\sqrt{255}\)

=1093+16

=1109