Question

 x^{4} -81-10x^{3}+90x=0

Answer 1

$x^{4} -81-10x^{3}+90x=0 \ (x^2-9)(x^2+9)-10x(x^2-9)=0 \ (x^2-9)(x^2+9-10x)=0 \ (x-3)(x+3)(x^2-10x+9)=0$

$1) \ x+3=0 \ x=-3 \ \ 2) \ x-3=0 \ x=3 \ \ 3) \ x^2-10x+9=0 \ x_1+x_2=10 cup x_1x_2=9 \ x_1=1 cup x_2=9$

Ответ: -3; 1; 3; 9