Помогите решить (2x-3)^2(x-3)=(2x-3)(x-3)^2 x^4=(x-20)^2 x^6=(x-5)^3

Решите задачу:

$(2x-3)^2(x-3)-(2x-3)(x-3)^2 =0\\ (2x-3)(x-3)(2x-3-x+3)=0\\ x(2x-3)(x-3)=0\\ \left [\begin{gathered}x=0\\x=1.5\\x=3\end{gathered}\right.$
$2) x^{4}=(x-20)^2\\ |x|^2=|x-20|\\ |x-20|=x^2$
$\\ \left [ \begin{gathered} \left \{ {x-20\geq0}\atop {x-20=x^2} }\right.\\\left\{{x-20<0}\atop {x-20=-x^2}}\right. \end{gathered}\right.$

$\left [ \begin{gathered} \left \{ {x\geq20}\atop {x^2-x+20=0} }\right.\\ \left \{ {x<20}\atop {x-20+x^2=0} }\right.\\ \end{gathered}\right.$

$\left [ \begin{gathered} \left \{ {x\geq20}\atop {\begin{gathered} x^2-x+20=0\\D=1-80<0;\ \Rightarrow\varnothing\end{gathered} }}\right. \\ \left \{ {x<20}\atop { \left[\begin{gathered}x=4\\x=-5\end{gathered}\right.} } \end{gathered}\right.\\ ANSWER: -5;4$

$\\\\3) x^6=(x-5)^3\\ (x^2)^{3}=(x-5)^3\\ x^2=x-5\\ x^2-x+5=0\\ D=1-20<0\Rightarrow\varnothing$