4В.38 4В.39 и 4В.40 пожалуйста

Решите задачу:

$x^3+x^2-9x-9=0 \\\ x^2(x+1)-9(x+1)=0 \\\ (x+1)(x^2-9)=0 \\\ (x+1)(x-3)(x+3)=0 \\\ x_1=-1; \ x_2=3; \ x_3=-3$

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

$3x^3-x^2+18x-6=0 \\\ 3x^2(x- \frac{1}{3}) +18(x- \frac{1}{3}) =0 \\\ (x- \frac{1}{3}) (3x^2+18) =0 \ [3x^2+18\ \textgreater \ 0] \\\ x- \frac{1}{3}=0\Rightarrow x= \frac{1}{3}$

$x^3-5x^2+9x-5=0 \\\ x^3-x^2-4x^2+4x+5x-5=0 \\\ x^2(x-1)-4x(x-1)+5(x-1)=0 \\\ (x-1)(x^2-4x+5)=0 \\\ (x-1)((x-2)^2+1)=0 \ [(x-2)^2+1\ \textgreater \ 0] \\\ x-1=0 \Rightarrow x=1$

$x^3-27x=0 \\\ x(x^2-27)=0 \\\ x(x-3 \sqrt{3} )(x+3 \sqrt{3} )=0 \\\ x_1=0 ; \ x_2=3 \sqrt{3} ; \ x_3=-3 \sqrt{3}$

$x^3-18x=0 \\\ x(x^2-18)=0 \\\ x(x-3 \sqrt{2} )(x+3 \sqrt{2} )=0 \\\ x_1=0 ; \ x_2=3 \sqrt{2} ; \ x_3=-3 \sqrt{2}$