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I saw a Youtube video in which it was shown that $$(7+50^{1/2})^{1/3}+(7-50^{1/2})^{1/3}=2$$ Since there are multiple values we can choose for the $3$rd root of a number, it would also make more sense to declare the value of this expression to be one of $2, 1 + \sqrt{-6},$ or $1 - \sqrt{-6}$

We may examine this more generally. If we declare $x$ such that $$x=(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}$$ $$\text{(supposing } a \text{ and } b \text{ to be integers here)}$$ one can show that $$x^3+3(b-a^2)^{1/3}x-2a=0$$ 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 $$x^3+3(b-a^2)^{1/3}x-2a$$ $$||$$ $$(x-n)(x^2+cx+d)$$ $$||$$ $$x^3+(c-n)x^2+(d-nc)x-nd$$ Since $(c-n)x^2=0$ we conclude that $c=n$ and we have $$x^3+3(b-a^2)^{1/3}x-2a=x^3+(d-c^2)x-cd$$ And - to continue our chain of conclusions - we conclude that $$3(b-a^2)^{1/3}=d-c^2 \quad\text{and}\quad 2a=cd$$ At this point I tried creating a single equation and got $$108b=4d^3+15c^2d^2+12c^4d-c^6$$ This is as far as I went.

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marked as duplicate by mercio, Lord Shark the Unknown, Michael Lugo, B. Goddard, Daniel W. Farlow Aug 24 '17 at 17:53

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  • $\begingroup$ See also: math.stackexchange.com/questions/386488/… $\endgroup$ – lab bhattacharjee Aug 24 '17 at 4:56
  • $\begingroup$ @JackD'Aurizio Consider the notation "$x^{1/3}$". We usually take this to mean a single number. But there is another sense in which "$x^{1/3}$" means any number $w$ such that $w\cdot w \cdot w=x$. That is to say, we are defining $x^{1/3}$ with a functional equation. And in this particular case , that functional equation is solved by $3$ distinct numbers when $x \ne 0$. In any case! - I amended my aforementioned language. $\endgroup$ – Christian Woll Aug 24 '17 at 5:43
  • $\begingroup$ @ChristianWoll: sorry for the nitpicking, good improvement by the way. $\endgroup$ – Jack D'Aurizio Aug 24 '17 at 13:17
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Use $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ identity. Things will cancel out.

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  • $\begingroup$ So the quantity $(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}$ is always an integer for any integer values of $a$ and $b$? $\endgroup$ – Jack D'Aurizio Aug 24 '17 at 2:03
  • $\begingroup$ It does not look so, and the question is asking for conditions granting that. $\endgroup$ – Jack D'Aurizio Aug 24 '17 at 2:04
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One choice is where $a-b^{1/2}=0$ so $a=4k^3$ for integer $k$ and $b=16k^6$, then $$(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}=2k$$

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