Consider the quartic
$-(a^5 + b^5)x^4 + (4a^5 - b^5)x^3 - (6a^5 + b^5)x^2 + (4a^5 + b^5)x - (a^5 + b^5) = 0$ where $a$ and $b$ are nonzero fixed integers.
By inspection, one can quickly find that $(a, b, x)= (a, -a, 0) $ is a trivial rational solution. But are there any other ?
Are you sure about the coefficient of $x$? All the other terms have $-b^5$; that is the only one with $b^5$.
$-(a^5 + b^5)x^4 + (4a^5 - b^5)x^3 - (6a^5 + b^5)x^2 + (4a^5 + b^5)x - (a^5 + b^5) = 0 $
Anyway, I will try to simplify it and see what happens.
$\begin{array}\\ -(a^5 + b^5)x^4 + (4a^5 - b^5)x^3 - (6a^5 + b^5)x^2\\ \quad + (4a^5 + b^5)x - (a^5 + b^5) &=-a^5x^4 + 4a^5x^3 - 6a^5x^2 + 4a^5x - a^5\\ &\quad-b^5x^4 - b^5x^3 -b^5x^2 + b^5x - b^5\\ &=-a^5(x^4 - 4x^3 + 6x^2 - 4x +1) -b^5(x^4 +x^3 +x^2 -x +1)\\ &=-a^5(x-1)^4 -b^5(x^4 +x^3 +x^2 -x +1)\\ &=-a^5(x-1)^4 -b^5(x^4 +x^3 +x^2 +x +1)+2b^5x\\ &=-a^5(x-1)^4 -b^5(x^5-1)/(x-1)+2b^5x\\ \end{array} $
Eh.
Wolfy gives four incredible complex expressions for the roots and $x=0$ and $a=-b$ and $b \ne 0$.
By the rational root theorem, any rational root $r = p/q$ must have $p | a^5+b^5$ and $q | a^5+b^5$. Since $a^5+b^5 =(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4) $, we can try $a+b$ and $1/(a+b)$. Neither seems to work.
That's all, folks.