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

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.
