Question

$x^{2}+\frac{x}{x-1} ^2=8$

Answer 1

$x^{2}+\frac{x}{x-1} ^2=8$

$x^{2}+(\frac{x}{x-1})^{2}=8,x\neq1$

$x^{2}+\frac{x^{2} }{(x-1)^{2} }=8$

$x^{2}+\frac{x^2}{(x-1)^2}-8=0$

$\frac{(x-1)^2*x^2+x^2-8(x-1)^2}{(x-1)^{2} } =0$

$\frac{((x-1)x)^{2}+x^2-8(x^2-2x+1)}{(x-1)^{2} } =0$

$\frac{x^4-2x^3+x^2+x^2-8x^2+16x-8}{(x-1)^2} = 0$

$\frac{x^4-2x^3-6x^2+16x-8}{(x-1)^2} = 0$

$x^{4}-2x^{3}-6x^{2}+16x-8=0$

$x^{4}-2x^{3}-6x^{2}+12x+4x-8=0$

$x^{3}*(x-2)-6x*(x-2)+4(x-2)=0$

$(x-2)*(x^3-6x+4)=0$

$(x-2)*(x^3-2x^2+2x^2-4x-2x+4)=0$

$(x-2)*(x^2*(x-2)+2x*(x-2)-2(x-2))=0$

$(x-2)*(x-2)*(x^2+2x-2)=0$

$(x-2)^{2}*(x^2+2x-2)=0$

$(x-2)^{2}=0$

$x^{2}+2x-2=0$

$x=2$

$x=-1+\sqrt{3},x\neq 1$

$x=-1-\sqrt{3}$

$x=2$

$x=-1+\sqrt{3}$

$x=-1-\sqrt{3}$

$x_1=-1-\sqrt{3}$

$x_2=-1+\sqrt{3}$

$x_{3}=2$