Just like in the title, I wonder if there is a way to solve $2x(1+\frac{1}{x^3})+7(1+\frac{1}{x})=0$ without any use of solving cubics including the rational root theorem (with these methods I already got $x=-2,-1,-\frac{1}{2}$).
The original equation is supposed to be related to quadratic substitution (the ones where you do $t=x^2$, like $x^4-2x^2+1=0$).
Thanks in advance for any help!
On
$$2x\left( 1+\frac { 1 }{ x^{ 3 } } \right) +7\left( 1+\frac { 1 }{ x } \right) =0\\ \left( 1+\frac { 1 }{ x } \right) \left( 2x\left( 1-\frac { 1 }{ x } +\frac { 1 }{ { x }^{ 2 } } \right) +7 \right) =0\\ \left( 1+\frac { 1 }{ x } \right) \left( 2x+5+\frac { 2 }{ { x } } \right) =0\\ 1+\frac { 1 }{ x } =0\Rightarrow x+1=0\Rightarrow x=-1\\ 2x+5+\frac { 2 }{ { x } } =0\\ 2{ x }^{ 2 }+5x+2=0\\ x=\frac { -5\pm 3 }{ 4 } \\ \\ \\ \\ \\ $$
Hint $$\left(1+\frac{1}{x^3}\right)=\left(1+\frac{1}{x}\right)\left(1-\frac{1}{x}+\frac{1}{x^2}\right)$$