Vtyjkyuk

Let $R$ denote a non-negative rational number. Determine a fixed set of integers $a,b,c,d,e,f$, such that for every choice of $R$,

$left|fracaR^2+bR+cdR^2+eR+f-sqrt[3]2right|<|R-sqrt[3]2|$

The given inequality seemed too much similar to limits so I applied R tending to $sqrt[3]2$. And then I tried solving the equation I get however I am unable to solve for integers. Any help please

(Too long for a comment.) The question is missing a lot of context, so it is hard to guess if the following even applies at all. But FWIW the given form reminisces of an iterated approximation steadily approaching $,sqrt[3]2,$.

Newton-Raphson would come to mind first, but the canonical $,f(x) = x^3-2,$ doesn't give $,x - f(x) / f'(x),$ in the prescribed form of a ratio of two quadratics.

Another way to "guess" an answer could be to start from the identity $,a^3-b^3=(a-b)(a^2+ab+b^2),$, so $,sqrt[3]2=1+1 big/ left(left(sqrt[3]2right)^2+sqrt[3]2+1right),$ i.e. $,sqrt[3]2$ is a fixed point of $,f(x)=(x^2+x+2)/(x^2+x+1),$ which corresponds to $,a=b=d=e=f=1, c=2,$ and does indeed appear to converge as required. Formally proving that convergence would take some more work, though.

The given inequality seemed too much similar to limits so I applied R tending to $sqrt[3]2$.

Formalizing the above, let $,f(x)=dfracax^2+bx+cdx^2+ex+f,$ and consider the sequence defined by $,r_n+1=f(r_n),$ for some $,r_0 in mathbbR,$. Furthermore, suppose that $,r_n to r = sqrt[3]2,$, then passing the recurrence relation to the limit gives:

$$r=f(r) ;;iff;; dr^3+(e-a)r^2+(f-b)r-c = 0$$

But the minimal polynomial of $,r=sqrt[3]2,$ over $,Bbb Z[X],$ is $x^3 - 2$, so it follows that $,beginalignbegincases2d &= c \ e&=a \f&=bendcasesendalign,$ and therefore $,f(x) = dfracax^2+bx+2ddx^2+ax+b,$ or, choosing WLOG $,d=1,$, $,f(x) = dfracax^2+bx+2x^2+ax+b,$ .

The case given in the original answer $,f(x)=dfracx^2+x+2x^2+x+1,$ corresponds to $,a=b=1,$ above.