## Jonask 3 years ago which one is larger

$\sqrt[3]{60}$ $2+\sqrt[3]{7}$no machines allowed

$2=\sqrt[3]{8}$

$60=(\frac{ 7+8 }{ 2 })10$

$\sqrt[3]{60}= \sqrt[3]{5}(\sqrt[3]{7+8})$

almost simlar now to$\sqrt[3]{8}+\sqrt[3]{7}$ HALT

!!!!!!

is comparing 60 with $(2+\sqrt[3]{7})^3$the same idea(tautology)

i think we have to use inequalities $(2+\sqrt[3]{7})^3=8+3(4)\sqrt[3]{7}+3(2)(\sqrt[3]{7})^2+7$

$\sqrt[3]{1}<\sqrt[3]{7}<\sqrt[3]{8}$

$15+12a+6a^2$ and 60

11. klimenkov

Interesting question!

$a=\sqrt[3]{7}$

we can allow machines

$a<2$

$\frac{ x+y+z }{ 3 } \leq \sqrt[3]{xyz}$ useful ineq

geometric mean

exhausted

18. klimenkov

Try to do like this. How to evaluate $$\sqrt[3]{7}$$. $$1<{7}<8 \Rightarrow 1<\sqrt[3]{7}<2$$. As 7 is closer to 8 than to 1, I hope $$\sqrt[3]{7}$$ is closer to 2. Lets put $$\sqrt[3]{7}=2+\alpha$$. Then $$7=8+12\alpha+6\alpha^2+\alpha^3$$. Since $$\alpha <1$$ we can neglect $$\alpha^2 ,\alpha^3$$. So, $$\alpha=-\frac1{12}$$. So the approximation for $$\sqrt[3]{7}$$ is $$2-\frac1{12}=\frac{23}{12}$$. Do this once again, because it is not a good accuracy for this problem.