ПОМОГИТЕ!!! ДАЮ 30 БАЛЛОВ

Ответ:

$\frac{x}{y} \boxed{\; a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)\; }\\\\\\x^3-8=(x-2)(x^2+2x+4)\\\\\\\dfrac{1}{27}+c^3=(\dfrac{1}{3}+c)(\dfrac{1}{9}-\dfrac{1}{3}\, c+c^2)\\\\\\0,008\, x^3-\dfrac{1}{64}\, y^3=\Big(0,2x-\dfrac{1}{4}\, y\Big)\Big(0,4x^2+0,05xy+y^2\Big)\\\\\\y^3+27=(y+3)(y^2-3y+9)\\\\\\\dfrac{1}{8}-x^3=\Big(\dfrac{1}{2}-x\Big)\Big(\dfrac{1}{4}+\dfrac{1}{2}\, x+x^2\Big)=(0,5-x)(0,25+0,5x+x^2)$

$\dfrac{1}{27}\, x^3+0,001d^3=\Big(\frac{1}{3}\, x+0,1d\Big)\Big(\dfrac{1}{9}\, x-\dfrac{1}{30}\, x\, d+0,01\, d^2\Big)\\\\\\c^3-1=(c-1)(c^2+c+1)\\\\\\\dfrac{1}{64}+y^3=\Big(\dfrac{1}{4}+y\Big)\Big(\dfrac{1}{16}-\dfrac{1}{4}\, y+y^2\Big)=(0,25+y)(0,0625-0,25\, y+y^2)\\\\\\\\0,125\, a^3-\dfrac{1}{8}\, b^3=\Big(\dfrac{1}{2}\, a-\dfrac{1}{2}\, b\Big)\Big(\dfrac{1}{4}\, a+\dfrac{1}{4}\, ab+\dfrac{1}{4}\, b^2\Big)=\\\\{}\ \ \ \ \ =(0,5\, a-0,5b)(025a+0,25ab+0,25b^2)=0,125(a-b)(a^2+ab+b^2)$

ПОМОГИТЕ!!! ДАЮ 30 БАЛЛОВ​

ПОМОГИТЕ!!! ДАЮ 30 БАЛЛОВ