Jonask
which one is larger
Delete
Share
This Question is Closed
Jonask
Best Response
You've already chosen the best response.
0
\[\sqrt[3]{60}\]
\[2+\sqrt[3]{7}\]no machines allowed
Jonask
Best Response
You've already chosen the best response.
0
\[2=\sqrt[3]{8}\]
Jonask
Best Response
You've already chosen the best response.
0
\[60=(\frac{ 7+8 }{ 2 })10\]
Jonask
Best Response
You've already chosen the best response.
0
\[\sqrt[3]{60}= \sqrt[3]{5}(\sqrt[3]{7+8})\]
Jonask
Best Response
You've already chosen the best response.
0
almost simlar now to\[\sqrt[3]{8}+\sqrt[3]{7}\]
HALT
Jonask
Best Response
You've already chosen the best response.
0
!!!!!!
Jonask
Best Response
You've already chosen the best response.
0
is comparing 60 with \[(2+\sqrt[3]{7})^3\]the same idea(tautology)
Jonask
Best Response
You've already chosen the best response.
0
i think we have to use inequalities
\[(2+\sqrt[3]{7})^3=8+3(4)\sqrt[3]{7}+3(2)(\sqrt[3]{7})^2+7\]
Jonask
Best Response
You've already chosen the best response.
0
\[\sqrt[3]{1}<\sqrt[3]{7}<\sqrt[3]{8}\]
Jonask
Best Response
You've already chosen the best response.
0
\[15+12a+6a^2\] and 60
klimenkov
Best Response
You've already chosen the best response.
1
Interesting question!
Jonask
Best Response
You've already chosen the best response.
0
\[a=\sqrt[3]{7}\]
Jonask
Best Response
You've already chosen the best response.
0
we can allow machines
Jonask
Best Response
You've already chosen the best response.
0
\[a<2\]
Jonask
Best Response
You've already chosen the best response.
0
\[\frac{ x+y+z }{ 3 } \leq \sqrt[3]{xyz}\]
useful ineq
Jonask
Best Response
You've already chosen the best response.
0
geometric mean
Jonask
Best Response
You've already chosen the best response.
0
exhausted
klimenkov
Best Response
You'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.