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Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt[3]{60}\] \[2+\sqrt[3]{7}\]no machines allowed
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[2=\sqrt[3]{8}\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[60=(\frac{ 7+8 }{ 2 })10\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt[3]{60}= \sqrt[3]{5}(\sqrt[3]{7+8})\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
almost simlar now to\[\sqrt[3]{8}+\sqrt[3]{7}\] HALT
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
is comparing 60 with \[(2+\sqrt[3]{7})^3\]the same idea(tautology)
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
i think we have to use inequalities \[(2+\sqrt[3]{7})^3=8+3(4)\sqrt[3]{7}+3(2)(\sqrt[3]{7})^2+7\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt[3]{1}<\sqrt[3]{7}<\sqrt[3]{8}\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[15+12a+6a^2\] and 60
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Interesting question!
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[a=\sqrt[3]{7}\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
we can allow machines
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ x+y+z }{ 3 } \leq \sqrt[3]{xyz}\] useful ineq
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
geometric mean
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago
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