Jinbei

Đặt : $m=x^{2}; n=y^{2}$.

Bài toán trở thành : Cho $P=\sqrt{m+\sqrt[3]{m^{2}n}}+\sqrt{n+\sqrt[3]{n^{2}m}}$. Chứng minh : $\sqrt[3]{P^{2}}=\sqrt[3]{m}+\sqrt[3]{n}$

Có : 

$P^{2}=m+\sqrt[3]{m^{2}n}+n+\sqrt[3]{mn^{2}}+2\sqrt{mn+m\sqrt[3]{m^{2}n}+n\sqrt[3]{n^{2}m}+mn}$

$\Rightarrow P^{2}=(\sqrt[3]{m}+\sqrt[3]{n})(\sqrt[3]{m^{2}}−\sqrt[3]{mn}+\sqrt[3]{n^{2}})+\sqrt[3]{mn}(\sqrt[3]{m}+\sqrt[3]{n})+2\sqrt{2.\sqrt[3]{m^{3}n^{3}}+\sqrt[3]{m^{2}n^{2}(\sqrt[3]{m}+\sqrt[3]{n})}}$

$\Rightarrow P^{2}=(\sqrt[3]{m}+\sqrt[3]{n})^{2}+2\sqrt[3]{mn}(\sqrt[3]{m}+\sqrt[3]{n})$  

(do : $\sqrt{\sqrt[3]{m^{2}n^{2}}}=\sqrt[3]{\sqrt{m^{2}n^{2}}}=\sqrt[3]{mn}$)

$\Rightarrow P^{2}=(\sqrt[3]{m}+\sqrt[3]{n})^{3}\Rightarrow \sqrt[3]{P^{2}}=\sqrt[3]{m}+\sqrt[3]{n}$ (đpcm)

Đặt :

$ax^{3}+by^{3}+cz^{3}=k^{3}\Rightarrow x\sqrt[3]{a}=y\sqrt[3]{b}=z\sqrt[3]{x}=k$

$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\sqrt[3]{a}}{k}+\frac{\sqrt[3]{b}}{k}+\frac{\sqrt[3]{c}}{k}=1$

$\Rightarrow \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=k$     $(1)$

Có : 

$\sqrt[3]{ax^{2}+by^{2}+cz^{2}}=\sqrt[3]{\frac{k^{3}}{x}+\frac{k^{3}}{y}+\frac{k^{3}}{z}}=k.\sqrt[3]{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=k$  $(2)$

Từ (1) và (2) suy ra đpcm.