(1+y^2)^2=4y(1+y)^2 ^2-в квадрате

Решите задачу:

$(1+y^2)^2=4y(1+y)^2;\\y^4+2y^2+1=4y(y^2+2y+1);\\y^4+2y^2+1=4y^3+8y^2+4y;\\y^4-4y^3-6y^2-4y+1=0 |:y^2\neq 0;\\y^2-4y-6-\frac{4}{y}+\frac{1}{y^2}=0;\\y^2+\frac{1}{y^2}-4(y+\frac{1}{y})-6=0;\\(y+\frac{1}{y})^2-2-4(y+\frac{1}{y})-6=0;\\(y+\frac{1}{y})^2-4(y+\frac{1}{y})-8=0.$

$t=y+\frac{1}{y}.\\t^2-4t-8=0;\\D_t/4=4+8=12;\\t_1=2-2\sqrt{3},\\t_2=2+2\sqrt{3}.$

$y+\frac{1}{y}=2-2\sqrt{3};\\f(y)=y+\frac{1}{y},\\E(f)=(-\infty; -2]U[2; +\infty];\\f(y)\neq2-2\sqrt{3}.$

$y+\frac{1}{y}=2+2\sqrt{3};\\ y^2-(2+2\sqrt{3})y+1=0;\\D_y/4=(1+\sqrt{3})^2-1=1+2\sqrt{3}+3-1=3+2\sqrt{3};\\y_1=1+\sqrt{3}+\sqrt{3+2\sqrt{3}};\\y_2=1+\sqrt{3}-\sqrt{3+2\sqrt{3}}.$