Answers

2014-08-28T22:15:12+08:00
 \frac{27x^3-27x^2+9x-1 }{27x^3-1} =\frac{(3x -1)^3}{ 3^3x^3-1} =\frac{(3x -1)^3}{ (3 x)^3-1^3}= \frac{(3x -1)^3}{ (3 x -1)((3x )^2+3x\cdot 1+1^2)}= \\\\=\frac{(3x -1)^3}{ (3 x -1)(9x ^2+3x +1 )}=\frac{(3x -1)^2}{  9x ^2+3x +1 } \\\\27x^3-1\neq 0\\ \\27x^{3}=1 \ \ | \ divide \ both \ sides\  by\  27 \\\\x^3=\frac{1}{27}   \\\\x=\sqrt[3]{\frac{1}{27}} =\frac{1}{3}\\\\D=R\setminus \left \{ \frac{1}{3} \right \}  




\frac{x^6-y^6}{x^4-y^4 } =\frac{(x^3)^2-(y^3)^{2}}{(x^2)^{2}-( y^2)^{2} }=\\\\\\= \frac{(x^3-y^3 )(x^3+y^3)}{(x^2-y^2)(x^2+y^2)}=\frac{(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)}{(x -y )(x+y)(x^2+y^2)}=\\\\\\=\frac{ (x^2+xy+y^2) (x^2-xy+y^2)}{  x^2+y^2 }=\frac{ x^4-x^3y +x^2y^2+x^3y-x^2y^2+xy^3+x^2y^2-xy^3+y^4 }{  x^2+y^2 }= \\\\\\=\frac{ x^4  +x^2y^2 +y^4 }{  x^2+y^ )}  \\\\x\neq y
1 5 1