I saw a Youtube video in which it was shown that
(7+501/2)1/3+(7−501/2)1/3=2
Since there are multiple values we can choose for the 3rd root of a number, it would also make more sense to declare the value of this expression to be one of 2,1+√−6, or 1−√−6
We may examine this more generally. If we declare x such that
x=(a+b1/2)1/3+(a−b1/2)1/3
(supposing a and b to be integers here)
one can show that
x3+3(b−a2)1/3x−2a=0
Which indeed has 3 roots.
We now ask
For what integer values of a and b is this polynomial solved by an integer?
I attempted this by assuming that n is a root of the polynomial. We then have
x3+3(b−a2)1/3x−2a
||
(x−n)(x2+cx+d)
||
x3+(c−n)x2+(d−nc)x−nd
Since (c−n)x2=0 we conclude that c=n and we have
x3+3(b−a2)1/3x−2a=x3+(d−c2)x−cd
And - to continue our chain of conclusions - we conclude that
3(b−a2)1/3=d−c2and2a=cd
At this point I tried creating a single equation and got
108b=4d3+15c2d2+12c4d−c6
This is as far as I went.
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