A good problem for a good year 2018

$x+ \dfrac 1{x+3} = -5$

$\Rightarrow x(x+3) + \dfrac 1{x+3} (x+3) = -5(x+3)$

$\Rightarrow x(x+3) + 1 = -5(x+3)$

$\Rightarrow x^2 + 3x + 1 = -5x - 15$

$\Rightarrow x^2 + 8x +16 = 0$

$\Rightarrow (x + 4)^2 = 0$

We get $x = -4$

Hence $(-4 +3)^{2018} + \dfrac 1{(-4 +3)^{2018}} = (-1)^{2018} + \dfrac{1}{(-1)^{2018}} =\boxed{2}$

You just need to manipulate it to get a perfect square format Once you solve the equations connected by the arrow , you get $x = -4$ . Then you input it back into the original equation. You get $2$ as the answer.