Answer
The solution to this equation is $x=-\dfrac{1}{2}$
Work Step by Step
$\dfrac{1}{x^{3}}+\dfrac{4}{x^{2}}+\dfrac{4}{x}=0$
Multiply the whole equation by $x^{3}$:
$x^{3}\Big(\dfrac{1}{x^{3}}+\dfrac{4}{x^{2}}+\dfrac{4}{x}=0\Big)$
$1+4x+4x^{2}=0$
Rearrange:
$4x^{2}+4x+1=0$
This is a perfect square trinomial, factor it:
$(2x+1)^{2}=0$
Take the square root of both sides:
$\sqrt{(2x+1)^{2}}=\sqrt{0}$
$2x+1=0$
Solve for $x$:
$2x=-1$
$x=-\dfrac{1}{2}$
The solution to this equation is $x=-\dfrac{1}{2}$
