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

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

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

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

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

JonaskBest 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

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

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

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

klimenkovBest 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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