Let $\sqrt[3]{\sqrt{50}+7}=a, \sqrt[3]{\sqrt{50}-7}=b$. Then $ab=1,a^3-b^3=14$. If $r=a-b$, then $a^3-b^3=r((a-b)^2+3ab)=r^3+3r$. But $r^3+3r=14$ had not rational solutions. Therefore you are wrong.

Rust wrote: But $r^3+3r=14$ had not rational solutions. Therefore you are wrong. $r=2$ $2^3+3(2)=8+6=14$

My solution:

master_Hjom wrote: My solution:

Wow, magic. How do u do that. I dont understand.

In fact, from the property of conjugate surds it is almost obvious. Isn't it? If $\sqrt[3]{\sqrt{50}+7}=x+\sqrt y,\sqrt[3]{\sqrt{50}-7}=x-\sqrt y$ Then the task is to find such $x,y$ and it is easy to note that $x=1,y=2$

MathSolver94 wrote: master_Hjom wrote: My solution:

Wow, magic. How do u do that. I dont understand. Anyway.. I used $(a+b)^3 = a^3 + 3a^2b+3ab^2+b^3$

mathmdmb wrote: If $\sqrt[3]{\sqrt{50}+7}=x+\sqrt y,\sqrt[3]{\sqrt{50}-7}=x-\sqrt y$ Then the task is to find such $x,y$ and it is easy to note that $x=1,y=2$ Do $x,y$ must be rational or real numbers? master_Hjom wrote: $\sqrt[3]{7+5\sqrt{2}}=\sqrt[3]{1^{3}+3\cdot 1^{2}\cdot\sqrt{2}+3\cdot 1\cdot\sqrt{2}^{2}+(2\sqrt{2})^{3}} $ How did u notice that?

Quote: master_Hjom wrote: $\sqrt[3]{7+5\sqrt{2}}=\sqrt[3]{1^{3}+3\cdot 1^{2}\cdot\sqrt{2}+3\cdot 1\cdot\sqrt{2}^{2}+(2\sqrt{2})^{3}} $ How did u notice that? Well, it's standard method for solving this radical stuffs... Also, tomorrow I'll have a math test in school which includes such things.. If you want to practise more things like this, here are you (from my school book): 1. $\sqrt{40\sqrt{2} - 57} - \sqrt{40\sqrt{2} + 57}$ 2. $\sqrt{8+2\sqrt{10+2\sqrt{5}}} + \sqrt{8-2\sqrt{10+2\sqrt{5}}}$ 3. $\sqrt[3]{9\sqrt{3} - 11\sqrt{2}}$ 4. $\sqrt[3]{2+\frac{10}{3\sqrt{3}}} + \sqrt[3]{2-\frac{10}{3\sqrt{3}}}$ 5. $\sqrt[3]{6+\sqrt{\frac{847}{27}}} + \sqrt[3]{6-\sqrt{\frac{847}{27}}}$

Let $x=\sqrt[3]{\sqrt{50}+7}-\sqrt[3]{\sqrt{50}-7}$, using $\boxed{\boxed{A+B+C=0\Longrightarrow A^3+B^3+C^3=3ABC}}$, $x^3-(50+\sqrt{7})+(\sqrt{50}-7)=3x(-\sqrt[3]{\sqrt{50}+7})(\sqrt[3]{\sqrt{50}-7})$ $\Longleftrightarrow x^3+3x-14=(x-2)(x^2+2x+7)=0\ (x>0)$, yielding $x=2=\frac 21.$

let $\frac{m}{n}=t$then $t^3+3t-14=0$ so$t=2$