Question

Cmr:

\(\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\dfrac{1}{9}}-\sqrt[3]{\dfrac{2}{9}}+\sqrt[3]{\dfrac{4}{9}}\)

Answer 1

Đặt \(\sqrt[3]{2}=x\Rightarrow2=x^3\Rightarrow x^3+1=3;x^3-1=1\)

\(\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{x-1}=\sqrt[3]{\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}}=\sqrt[3]{\dfrac{x^3-1}{x^2+x+1}}\)

\(=\sqrt[3]{\dfrac{1}{x^2+x+1}}=\sqrt[3]{\dfrac{1}{x^2+x+\dfrac{1}{3}\left(x^3+1\right)}}\)

\(=\sqrt[3]{\dfrac{3}{x^3+3x^2+3x+1}}=\sqrt[3]{\dfrac{27}{9\left(x+1\right)^3}}=\dfrac{1}{\sqrt[3]{9}}.\dfrac{3}{x+1}\)

\(=\dfrac{1}{\sqrt[3]{9}}\left(\dfrac{x^3+1}{x+1}\right)=\dfrac{1}{\sqrt[3]{9}}\left(1-x+x^2\right)=\dfrac{1}{\sqrt[3]{9}}\left(1-\sqrt[3]{2}+\sqrt[3]{4}\right)\)

\(=\sqrt[3]{\dfrac{1}{9}}-\sqrt[3]{\dfrac{2}{9}}+\sqrt[3]{\dfrac{4}{9}}\) (đpcm)