Решите) с 1 по 4 зарание спасибо

Решите задачу:

$1.1) a^3+8b^3=(a+2b)(a^2+2ab+4b^2)\\1.2) x^2y-36y^3=y(x^2-36y^2)=y(x-6y)(x+6y)\\1.3) -5m^2+10mn-5n^2=-5(m-n)^2\\1.4) 4ab -28b+8a-56=4b(a-7)+8(a-7)=4(b+2)(a-7)\\1.5) a^4-81=(a^2-9)(a^2+9)=(a^2+9)(a-3)(a+3)\\2) a(a+2)(a-2)-(a-3)(a^2+3a+9)=a(a^2-4)-(a^3-27)=27-4a\\ 3.1) x-3y+x^2-9y^2=x-3y+(x-3y)(x+3y)=(x-3y)(x+3y+1)\\3.2) 9m^2+6mn+n^2-25=(3m+n)^2-25=(3m+n-5)(3m+n+5)\\3.3) ab^5-b^5-ab^3+b^3=b^5(a-1)-b^3(a-1)=(a-1)(b^5-b^3)=b^3(b-1)(b+1)(a-1)\\3.4) 1-x^2+10xy-25y^2=1-(x-5y)^2=(1-x+5y)(1+x-5y)\\$

$4.1) 3x^3-12x=0\\3x(x^2-4)=0\\3x(x-2)(x+2)=0\\x \in \{-2,0,2\}\\4.2) 49x^3+14x^2+x=0\\x(7x+1)^2=0\\x \in \{-\frac{1}{7},0\}\\4.3) x^3-5x^2-x+5=0\\x^2(x-5)-(x-5)=0\\(x-1)(x+1)(x-5)=0\\x \in \{-1,1,5\}$