How prove that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $

Question

How check that $ \sqrt[3]{\frac{1}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}=\sqrt[3]{\sqrt[3]2-1} $?

Answer 1

If we abbreviate $w=\sqrt[3]2$, the left hand side is $L=\frac1{\sqrt[3]9}(1-w+w^2)$. From $(1+w)(1-w+w^2)=1+w^3=3$ we see that $L=\frac3{\sqrt[3]9(1+\sqrt[3]2)}$, hence $$ L^3=\frac{27}{9(1+w)^3}=\frac3{1+3w+3w^2+w^3}=\frac1{1+w+w^2}$$ As above, note that $(1+w+w^2)(w-1)=w^3-1=1$, hence $$ L^3=w-1=R^3.$$

Answer 2

use three equations: $$a^3+b^3=(a+b)(a^2-ab+b^2)\quad (1)$$ $$a^3-b^3=(a-b)(a^2+ab+b^2)\quad (2)$$ $$(a+b)^3=a^3+3a^2b+3ab^2+b^3\quad (3)$$ for your problem: $$left\\=(\sqrt[3]{\frac{1}{3}})^2-\sqrt[3]{\frac{1}{3}}\sqrt[3]{\frac{2}{3}}+(\sqrt[3]{\frac{2}{3}})^2\\=\frac{\frac{1}{3}+\frac{2}{3}}{\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}}}\quad using(1)\\=\sqrt[3]{\frac{1}{(\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}})^3}}\\=\frac{\sqrt[3]{3}}{1+\sqrt[3]{2}}\\=\sqrt[3]{\frac{3}{(1+\sqrt[3]{2})^3}}\\=\sqrt[3]{\frac{3}{1+3\sqrt[3]{2}+3\sqrt[3]{2^2}+2}}\quad using(3)\\=\sqrt[3]{\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{2^2}}}\\=\sqrt[3]{\sqrt[3]{2}-1}\quad using(2)$$