Mathematical Reflections 2016, Issue 6 - Problem J331

Problem:
Solve the equation $$4x^3+\dfrac{127}{x}=2016.$$

Proposed by Adrian Andreescu, Dallas, Texas

Solution:
The given equation is equivalent to $$4x^4-2016x+127=0.$$ Observe that
$$\begin{array}{lll} 4x^4-2016x+127&=&4x^4-256x^2+(2x^2+16x)+127(2x^2-16x)+127\\&=&(2x^2-16x+1)(2x^2+16x)+127(2x^2-16x+1) \\ &=&(2x^2-16x+1)(2x^2+16x+127)\end{array}$$
So, the given equation becomes $$(2x^2-16x+1)(2x^2+16x+127)=0.$$
We obtain $$x=\dfrac{8 \pm \sqrt{62}}{2}, \qquad x=\dfrac{-8 \pm i\sqrt{190}}{2}.$$