Question

Решите пж, даю много баллов​

Answer 1

$\displaystyle\bf\\1)\\\\x^{2} -y^{2} -6x+9=(x^{2} -6x+9)-y^{2} =(x-3)^{2} -y^{2} =\\\\=(x-3-y)(x-3+y)\\\\\\2)\\\\x^{3} -3x^{2} -4x+12=0\\\\(x^{3} -3x^{2} )-(4x-12)=0\\\\x^{2} (x-3)-4(x-3)=0\\\\(x-3)(x^{2} -4)=0\\\\(x-3)(x-2)(x+2)=0\\\\x-3=0 \ \ ; \ \ x_{1} =3\\\\x-2=0 \ \ ; \ \ x_{2} =2\\\\x+2=0 \ \ ; \ \ x_{3} =-2\\\\Otvet:-2 \ ; \ 2 \ ; \ 3$