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If $a,b,c$ are rational numbers and if $\displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3$ is also rational then prove that $ab+bc+ca=0$

My attempt
Binomial expansion is not a good idea because there will be $27$ terms so I tried to prove using factorization.
$$\displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3-\left(c\sqrt[3]{4}\right)^3 \\=\left(a+ b\sqrt[3]{2}\right)\left[\left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^2+\left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)\left(c\sqrt[3]{4}\right)+\left(c\sqrt[3]{4}\right)^2\right]$$

This again leads to complicated calculations. Then I tried to equate it to a rational number $r$.

\begin{align*} \displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3&=r\\ \implies a+b\sqrt[3]{2}+c\sqrt[3]{4}&=r^{1/3}\\ \implies b\sqrt[3]{2}+c\sqrt[3]{4}&=r^{1/3}-a\\ \implies(b\sqrt[3]{2}+c\sqrt[3]{4})^3&=(r^{1/3}-a)^3\\ \implies2b^3+6\sqrt[3]{2}~b^2c+6\sqrt[3]{4}~bc^2+4c^3&=r-3r^{2/3}a+3r^{1/3}a^2-a^3 \end{align*} When I got stuck here I wrote the equation $a+b\sqrt[3]{2}+c\sqrt[3]{4}=r^{1/3}$ in $3$ different ways, each time multiplying with $\sqrt[3]{2}$ \begin{align*} \displaystyle a+b\sqrt[3]{2}+c\sqrt[3]{4}&=r^{1/3}\\ a\sqrt[3]{2}+b\sqrt[3]{4}+2c&=\sqrt[3]{2}r^{1/3}\\ a\sqrt[3]{4}+2b+2\sqrt[3]{2}c&=\sqrt[3]{4}r^{1/3}\end{align*} I tried adding the above three equations but it wasn't helpful.

Can someone help me in solving the question. Thanks in advance.

Is it possible to generalize the question as $\left(a+b\sqrt[3]{n}+c\sqrt[3]{n^2}\right)^3$ where $n$ is a non-square integer?

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    $\begingroup$ i think it would be tedious, but when you multiply this out, many of the radicals cancel and you can collect like terms fairly nicely $\endgroup$
    – C Squared
    Jul 7, 2021 at 8:25
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    $\begingroup$ So that the binomial expansion hurts less, I'd call $q = 2^{1/3}$ and simplify any exponent greater than $4$ using that $q^4 = 2q$. You can also ignore all rational terms, so you end up only with an equation with six terms, three with $q$ and three with $q^2$. $\endgroup$
    – AnilCh
    Jul 7, 2021 at 8:34
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    $\begingroup$ @CSquared I got this question from a book for contest maths, so it's more likely that there will be a solution without expanding the expression. $\endgroup$
    – Asher2211
    Jul 7, 2021 at 8:35
  • $\begingroup$ you could try to look for terms in the expansion which have one factors of $ac$, $bc$, or $ab$ in them and try to deduce something indirectly. $\endgroup$
    – C Squared
    Jul 7, 2021 at 8:40
  • $\begingroup$ Note: In your factorization, the second term should have a minus sign. $\endgroup$
    – Calvin Lin
    Jul 7, 2021 at 15:04

1 Answer 1

Reset to default
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By using $$(x+y+z)^3=x^3+y^3+z^3+3xy(x+y)+3xz(x+z)+3yz(y+z)+6xyz$$ we get $a^2b+2b^2c+2ac^2=0$ and $ab^2+a^2c+2bc^2=0$. Now multiply the first equation by $b$ and the second by $a$ and then subtract them. We obtain $c=0$ (and then $a=0$ or $b=0$) or $a=b=0$.

Edit. I forgot to mention that I used the following: if $p+q\sqrt[3]{2}+r\sqrt[3]{4}=0$, with $p,q,r$ rational numbers, then $p=q=r=0$.

The generalization is straightforward.

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    $\begingroup$ It seems to me that the goal is to prove $ab+bc+ca=0$ without proving that two among $a,b,c$ are zero. $\endgroup$
    – lhf
    Jul 7, 2021 at 16:09
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    $\begingroup$ @lhf You can give it a try. $\endgroup$
    – user26857
    Jul 7, 2021 at 17:31

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