Помогите, даю 50баллов

Решите задачу:

$1)\; \; -3,5ab^3c^2\cdot 1,6a^3bc=-5,6a^4b^4c^3\\\\(-2\frac{3}{4})b^4c^2\cdot (-\frac{8}{33})b^2c^4=\frac{11}{4}\cdot \frac{8}{33}b^6c^6=\frac{2}{3}b^6c^6\\\\2)\; \; (x-1)(x-2)(x+3)-(x+1)(x+2)(x-3)=\\\\=(x^2-3x+2)(x+3)-(x^2+3x+2)(x-3)=\\\\=x^3+3x^2-3x^2-9x+2x+6-(x^3-3x^2+3x^2-9x+2x-6)=\\\\=6+6=12$

$3)\; \; (2b+a^3)(a^3-2b)=(a^3)^3-(2b)^2=a^9-4b^2\\\\(x^2+y^2)(x^4-x^2y^2+y^4)=(x^2)^3+(y^2)^3=x^6+y^6\\\\4)\; \; 16ab^3-20a^2b^2=4ab^2\cdot (4b-5a)\\\\18x^4y^2-12x^5y^3=6x^4y^2\cdot (3-2xy)\\\\mn-2m+4n-8=m(n-2)+4(n-2)=(n-2)(m+4)\\\\5)\; \; (x-1)(\underbrace {x^7+x^6+x^5+x^4+x^3+x^2+x+1}_{geom.progr.\; ,\; b_1=1\; ,\; q=x})=\\\\=(x-1)\cdot \frac{1-b_{n}q}{1-q}=(x-1)\cdot \frac{1-x^7\cdot x}{1-x}=-(1-x^8)=x^8-1\; ;\\\\x^8-1=x^8-1$