Answer:
[tex]\huge\boxed{x=-2\ \vee\ x=-1\ \vee\ x=1}[/tex]
Step-by-step explanation:
[tex](x+1)^2-2=\dfrac{2}{x}\\\\\text{Domain:}\ x\neq0\\\\\dfrac{(x+1)^2-2}{1}=\dfrac{2}{x}\qquad\text{cross multiply}\\\\x\bigg((x+1)^2-2\bigg)=(1)(2)\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\x\bigg(x^2+(2)(x)(1)+1^2-2\bigg)=2\\\\x\bigg(x^2+2x+1-2\bigg)=2\\\\x\bigg(x^2+2x-1\bigg)=2\qquad\text{use the distributive property}\\\\(x)(x^2)+(x)(2x)+(x)(-1)=2[/tex]
[tex]x^3+2x^2-x=2\qquad\text{subtract 2 from both sides}\\\\x^3+2x^2-x-2=0\qquad\text{distributive}\\\\x^2(x+2)-1(x+2)=0\qquad\text{distributive}\\\\(x+2)(x^2-1)=0[/tex]
[tex]\text{The product is equal 0 if one of factors is equal 0.}\\\text{Therefore}\\\\(x+2)(x^2-1)=0\iff x+2=0\ \vee\ x^2-1=0\\\\x+2=0\qquad\text{subtract 2 from both sides}\\\\\boxed{x=-2}\in\text{Domain}\\\\x^2-1=0\qquad\text{add 1 to both sides}\\\\x^2=1\Rightarrow x=\pm\sqrt1\\\\\boxed{x=-1}\in\text{Domain}\ \vee\ \boxed{x=1}\in\text{Domain}[/tex]
