show this sequence always is rational number let $\{a_{n}\}$ such $a_{1}=-8$,and such
$$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$
show that
$$a_{n}\in Q,\forall n\in N^{+}$$
I  try let $a_{2}=x$,and for $n=1$, then we have
$$-8+5\sqrt[3]{x}=3\sqrt[3]{-49(x+1)}\Longrightarrow x=-1/8$$
and for $n=2$ I get $a_{3}=-\dfrac{389017}{4913}$ and so on
The first observation
i want use this well known identity: $$a^3+b^3+c^3=3abc ~~~~~~~~~~~~~~~~~~if~~~~~ a+b+c=0$$
let $a=4\sqrt[3]{a_{n}},b=5\sqrt[3]{a_{n+1}},c=-3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$,so
$$64a_{n}+125a_{n+1}-189(a_{n}+1)(a_{n+1}+1)=-180\sqrt[3]{7a_{n}a_{n+1}(a_{n}+1)(a_{n+1}+1)}$$
ADD it by 2021,11.6.PM.18:05
The second observation
and Now I have found this interesting:
if $x,y,p,q,a,b\ge 0$, Hölder's inequality :
$$(x^3+y^3)(p^3+q^3)(a^3+b^3)\ge (xpa+yqb)^3$$
$1+\sqrt[3]{e^{2a}}\sqrt[5]{e^{b}}\sqrt[15]{e^{2c}} \leq \sqrt[3]{(1+e^{a})^2}\sqrt[5]{1+e^{b}}\sqrt[15]{(1+e^{c})^2}$
the condition it's
$$(a_{n}+1)(1+a_{n+1})(4^3+5^3)= (4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}})^3$$
or
$$4\sqrt[3]{a_{n}}+5\sqrt[3]{a_{n+1}}=3\sqrt[3]{7(a_{n}+1)(a_{n+1}+1)}$$
but this sequence $a_{n}<0$ or $a_{n}>0$,so this Hölder inequality seem can't hold
 A: By looking at the initial values, we hypothesize that

*

*$a_n  = x_n^3, x_n \in \mathbb{Q}$

*$a_n + 1 = 7y_n^3, y_n \in \mathbb{Q}$
This suggests to study rational points on $7y^3 = x^3 + 1$.
We map $ a_i$ to the point $ ( \sqrt[3]{a_i}, \sqrt[3]{\frac{a_i + 1}{7} } )$.
We have the starting points $a_1: (-2, -1), a_2: (-1/2, 1/2)$.
Notice that we also have the rational points $ (-1, 0), (4/5, 3/5)$ on the curve. The latter was inspired by the coefficients in the question, and you will see how this is applicable in a moment.
We apply the usual tricks with group addition to determine the sequence.
Given a solution $(x,y)$, divide throughout by $x^3$ to get $ 7 (\frac{y}{x} ) ^3 = ( \frac{1}{x} ) ^3 + 1 $, so observe that $(\frac{1}{x}, \frac{y}{x} )$ is also a solution.
Given a solution $(x,y)$, we can construct the line that passes through $(x,y)$ and $(\frac{4}{5}, \frac{3}{5} )$ which intersects the cubic again at $(x', y')$.
Furthermore, if $(x,y)$ is a rational point, then so is $(x', y')$.  (Prove this.)
We combine this with the previous observation and send $(x,y)$ (to $(x',y')$ and then ) to $(\frac{1}{x'} , \frac{y'}{x'} ) $.
Claim: $(x,y)$ and $(\frac{1}{x'} , \frac{y'}{x'} )$ satisfy $4x + \frac{5}{x'} = 21 y \frac{y'}{x'}$.
See Zhaohui's solution for a beautiful proof that if the cubic intersects a line at 3 points, then $x_1x_2x_3 + 1  = 7 y_1y_2y_3$.
Applying it to the 3 points on the line, we get $\frac{4}{5} x x' + 1 = 7 \times \frac{3}{5} y y'$, and hence the claim follows.
Corollary: This shows that we send $ a_n : (x_n, y_n) $  to $a_{n+1} : (x_{n+1}, y_{n+1}) $ via the above description.
Since the starting point $a_1$ is rational, hence all of these are rational points.
So $ x_n \in \mathbb{Q}$ and $ a_n \in \mathbb{Q}$.
Notes

*

*Someone who is more familiar with elliptic curves might be able to manipulate the equations better. (IE See Zhaohui's solution)

*This approaches illustrates the theory behind the question, and why the question isn't just "magic".

*If one knew the fact that Zhaohui showed, then this approach would have felt natural. My guess is that similar questions like this can be dealt with in a similar way, especially for quadratic (and cubic) terms. Previously, I mainly dealt with similar questions previously via heavy-handed induction + guessing what the sequence is (EG Zhaohui's formulas at the end). I'm excited to apply this to future questions.

*For $a_1$, the line is actually tangential at $a_1$. So the 3rd point of intersection is $a_1$ (which explain why $a_1, a_2$ are related via $ (\frac{1}{x}, \frac{y}{x} )$.


Uncompleted attempt at proof: The line that passes through $(x,y)$ and $(\frac{4}{5}, \frac{3}{5} )$ is $ (Y' - y) ( \frac{4}{5} - x )  = (X' - x) ( \frac{3}{5} - y ) $, or that $Y' = X'\frac{5y-3}{5x-4} + \frac{3x-4y}{5x-4} $.
Substituting this into $7Y^3 = X^3 + 1$, and applying vieta's formula to find the sum of roots, we get that $\frac{4}{5} + x + x' = \frac{3 (\frac{5y-3}{5x-4})^2 (\frac{3x-4y}{5x-4}) } {1 - (\frac{5y-3}{5x-4})^3 } $, so $ x' = -\frac{625 x^4 - 1000 x^3 - 625 x y^3 + 675 x y + 370 x + 1000 y^3 - 900 y^2 - 148}{5 (5 x - 5 y - 1) (25 x^2 + 25 x y - 55 x + 25 y^2 - 50 y + 37))}$.
I'm unwilling to continue this tedious calculations, but one can find $y'$ and then verify that the equation holds.
A: This isn't an answer, really more of a musing on the problem, a full expression for $a_{n+1}$ in terms of $a_n$, and a way forward. Assuming that this problem is well posed (that is, $a_n$ can always be calculated from $a_{n-1}$ without ambiguity) then to get this expression, define $a_n=b_n^3$. Then the formula relating the $n$th and $(n+1)$th members (after cubing both sides) becomes
$$0=(4 b_n + 5 b_{n+1})^3 - 27 (7 (b_n^3 + 1) (b_{n_+1}^3 + 1))$$
$$=-189 b_{n+1}^3 b_n^3-125 b_n^3+240 b_{n+1} b_n^2+300 b_{n+1}^2 b_n-64 b_{n+1}^3-189$$
Since we are assuming that this cubic in $b_{n+1}$ has one real solution (otherwise the problem would be ill-posed) we can solve for $b_{n+1}$ in terms of $b_n$:
$$b_{n+1}=3 \sqrt[3]{\frac{7}{2}} \left(\sqrt[3]{-\frac{125 b_n^6+61 b_n^3-64}{\left(189
   b_n^3+64\right){}^2}-\frac{23625 b_n^9+27721 b_n^6+8192 b_n^3+4096}{\left(189
   b_n^3+64\right){}^3}}+\sqrt[3]{\frac{125 b_n^6+61 b_n^3-64}{\left(189
   b_n^3+64\right){}^2}-\frac{23625 b_n^9+27721 b_n^6+8192 b_n^3+4096}{\left(189
   b_n^3+64\right){}^3}}\right)-\frac{100 b_n}{-189 b_n^3-64}$$
We then use the above expression to get an expression for $a_{n+1}$ which is
$$a_{n+1}=\frac{\left(100 \sqrt[3]{a_n}-45\  \sqrt[3]{49a_n^2 \left(a_n+1\right)}-48
    \sqrt[3]{7\left(a_n+1\right){}^2}\right){}^3}{\left(189
   a_n+64\right){}^3}$$
One method to go forward here is to assume that $a_n=r^3$, and then it is sufficient to show that
$$a_n+1=7q^3$$
(for some rational $q$). This is because if this is the case then $a_{n+1}$ can also be written as $s^3$ for some rational $s$. Of course, it is not enough to simply assume $a_n=r^3$ because there are many rational $r$ for which the above equation does not hold (for example $r=1$). It is only for some $r$ (for example $r=-2$) where the above expression evaluates correctly and showing that this will always be the case for $a_n$ seems to be the difficult thing.
