[Math] Calculate $S=3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}-\sqrt[3]{2}-\sqrt[3]{20}+\sqrt[3]{25}$

Question

Calculate $$S=3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}-\sqrt[3]{2}-\sqrt[3]{20}+\sqrt[3]{25}$$ $\color{red}{\text{without using calculator}.}$

Please help me, I can't find any solution to sovle it.

Answer 1

Let

$$s=\sqrt[3]2+\sqrt[3]{20}-\sqrt[3]{25}$$

If you carefully square $s$, you find

$$s^2=9(\sqrt[3]5-\sqrt[3]4)$$

Note that $\sqrt[3]2\gt1$, $\sqrt[3]{20}\gt2$, and $\sqrt[3]{25}\lt3$, so $s\gt0$. Thus $s=3\sqrt{\sqrt[3]5-\sqrt[3]4}$, hence $S=0$.