# Exercise ID259

Algebra → Exponents → Powers
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 10] [Number of characters: 0]

Evaluate the following expression:

$\frac{\frac{1}{x}-64}{4+\frac{2}{x}+\frac{1}{{x}^{2}}}·\frac{{x}^{2}}{4-\frac{4}{x}+\frac{1}{{x}^{2}}}-\frac{4{x}^{2}\left(2x+1\right)}{1-2x}x\in Z$

$\frac{\frac{1}{x}-64}{4+\frac{2}{x}+\frac{1}{{x}^{2}}}·\frac{{x}^{2}}{4-\frac{4}{x}+\frac{1}{{x}^{2}}}-\frac{4{x}^{2}\left(2x+1\right)}{1-2x}$ $=\frac{\frac{1-64{x}^{6}}{{x}^{6}}}{\frac{4{x}^{2}+2x+1}{{x}^{2}}}·\frac{{x}^{2}}{\frac{4{x}^{2}-4x+1}{{x}^{2}}}$ $=\frac{1-64{x}^{6}}{{{x}}^{4}\left(4x+2x+1\right)}·\frac{{{x}}^{4}}{4{x}^{4}-4x+1}$ $=\frac{1-{\left(8{x}^{3}\right)}^{2}}{\left(4{x}^{2}+2x+1\right)\left(4{x}^{2}-4+1\right)}$ $=\frac{\left(1-8{x}^{3}\right)\left(1+8{x}^{3}\right)}{\left(4{x}^{2}+2x+1\right){\left(1-2x\right)}^{2}}$ $=\frac{\left(1-2x\right)\left(1+2x+4{x}^{2}\right)\left(1+2x\right)\left(1-2x+4{x}^{2}\right)}{\left(4{x}^{2}+2x+1\right){\left(1-2x\right)}^{2}}$ $=\frac{\left(1+2x\right)\left(1-2x+4{x}^{2}\right)}{\left(1-2x\right)}-\frac{4{x}^{2}\left(2x+1\right)}{\left(1-2x\right)}$ $=\frac{\left(2x+1\right)\left(4{x}^{2}-2x+1-4{x}^{2}\right)}{\left(1-2x\right)}$ $=\frac{\left(2x+1\right)\left(1-2x\right)}{{1}{-}{2}{x}}$ $=\left(2x+1\right)$
$\frac{\frac{1}{x}-64}{4+\frac{2}{x}+\frac{1}{{x}^{2}}}·\frac{{x}^{2}}{4-\frac{4}{x}+\frac{1}{{x}^{2}}}-\frac{4{x}^{2}\left(2x+1\right)}{1-2x}=\left(2x+1\right)$