Math Trivia -- A sum of cube roots of certain consecutive positive integers

lookagain

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Aug 22, 2010
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There is nothing here for which to solve. I was experimenting, and I discovered that the following sum
is relatively close to an integer:

\(\displaystyle \sqrt[3]{2} \ + \ \sqrt[3]{3} \ + \ \sqrt[3]{4} \ + \ \sqrt[3]{5} \ \approx \ 6 \)


The sum is approximately equal to 5.9995476.
 
There is nothing here for which to solve. I was experimenting, and I discovered that the following sum
is relatively close to an integer:

\(\displaystyle \sqrt[3]{2} \ + \ \sqrt[3]{3} \ + \ \sqrt[3]{4} \ + \ \sqrt[3]{5} \ \approx \ 6 \)


The sum is approximately equal to 5.9995476.
For more similar facts, see



I don't find yours there (though I didn't scan closely). The important thing is, people do find these interesting!
 
This reminds me of this question (which can be solved algebraically).
[math]\sqrt[3]{x} + \sqrt[3]{x+1}+ \sqrt[3]{x+2} = 6[/math]
There are 1 real, and 2 complex roots.
 
For more similar facts, see



I don't find yours there (though I didn't scan closely). The important thing is, people do find these interesting!
The one relating pi and the golden ratio is still a prominent on-going conspiracy.
 
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