Solve x^-6-64/4+2x^-1+x^-2x^2/4-4{x^-1+{x}^-2}-frac{4{x}^2left(2x+1right)}{1-2x}

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

Evaluate

$$2x+1$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{x^{-6}-64}{4+2x^{-1}+x^{-2}}.

$$\frac{-x^{-6}\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)\left(4x^{2}+2x+1\right)}{x^{-2}\left(4x^{2}+2x+1\right)}\times \frac{x^{2}}{4-4x^{-1}+x^{-2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

Cancel out 4x^{2}+2x+1 in both numerator and denominator.

$$\frac{-x^{-6}\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{-2}}\times \frac{x^{2}}{4-4x^{-1}+x^{-2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{-\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{4}}\times \frac{x^{2}}{4-4x^{-1}+x^{-2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

Factor the expressions that are not already factored in \frac{x^{2}}{4-4x^{-1}+x^{-2}}.

$$\frac{-\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{4}}\times \frac{x^{2}}{x^{-2}\left(2x-1\right)^{2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent.

$$\frac{-\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{4}}\times \frac{x^{4}}{\left(2x-1\right)^{2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

Multiply \frac{-\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{4}} times \frac{x^{4}}{\left(2x-1\right)^{2}} by multiplying numerator times numerator and denominator times denominator.

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

Cancel out \left(2x-1\right)x^{4} in both numerator and denominator.

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

Use the distributive property to multiply 4x^{2} by 2x+1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x-1 and 1-2x is 2x-1. Multiply \frac{8x^{3}+4x^{2}}{1-2x} times \frac{-1}{-1}.

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

Since \frac{-\left(2x+1\right)\left(4x^{2}-2x+1\right)}{2x-1} and \frac{-\left(8x^{3}+4x^{2}\right)}{2x-1} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in -\left(2x+1\right)\left(4x^{2}-2x+1\right)-\left(-\left(8x^{3}+4x^{2}\right)\right).

$$\frac{-8x^{3}+4x^{2}-2x-4x^{2}+2x-1+8x^{3}+4x^{2}}{2x-1}$$

Combine like terms in -8x^{3}+4x^{2}-2x-4x^{2}+2x-1+8x^{3}+4x^{2}.

$$\frac{4x^{2}-1}{2x-1}$$

Factor the expressions that are not already factored in \frac{4x^{2}-1}{2x-1}.

$$\frac{\left(2x-1\right)\left(2x+1\right)}{2x-1}$$

Cancel out 2x-1 in both numerator and denominator.

$$2x+1$$

Expand

$$2x+1$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{x^{-6}-64}{4+2x^{-1}+x^{-2}}.

$$\frac{-x^{-6}\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)\left(4x^{2}+2x+1\right)}{x^{-2}\left(4x^{2}+2x+1\right)}\times \frac{x^{2}}{4-4x^{-1}+x^{-2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

Cancel out 4x^{2}+2x+1 in both numerator and denominator.

$$\frac{-x^{-6}\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{-2}}\times \frac{x^{2}}{4-4x^{-1}+x^{-2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{-\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{4}}\times \frac{x^{2}}{4-4x^{-1}+x^{-2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

Factor the expressions that are not already factored in \frac{x^{2}}{4-4x^{-1}+x^{-2}}.

$$\frac{-\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{4}}\times \frac{x^{2}}{x^{-2}\left(2x-1\right)^{2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent.

$$\frac{-\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{4}}\times \frac{x^{4}}{\left(2x-1\right)^{2}}-\frac{4x^{2}\left(2x+1\right)}{1-2x}$$

Multiply \frac{-\left(2x-1\right)\left(2x+1\right)\left(4x^{2}-2x+1\right)}{x^{4}} times \frac{x^{4}}{\left(2x-1\right)^{2}} by multiplying numerator times numerator and denominator times denominator.

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

Cancel out \left(2x-1\right)x^{4} in both numerator and denominator.

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

Use the distributive property to multiply 4x^{2} by 2x+1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x-1 and 1-2x is 2x-1. Multiply \frac{8x^{3}+4x^{2}}{1-2x} times \frac{-1}{-1}.

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

Since \frac{-\left(2x+1\right)\left(4x^{2}-2x+1\right)}{2x-1} and \frac{-\left(8x^{3}+4x^{2}\right)}{2x-1} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in -\left(2x+1\right)\left(4x^{2}-2x+1\right)-\left(-\left(8x^{3}+4x^{2}\right)\right).

$$\frac{-8x^{3}+4x^{2}-2x-4x^{2}+2x-1+8x^{3}+4x^{2}}{2x-1}$$

Combine like terms in -8x^{3}+4x^{2}-2x-4x^{2}+2x-1+8x^{3}+4x^{2}.

$$\frac{4x^{2}-1}{2x-1}$$

Factor the expressions that are not already factored in \frac{4x^{2}-1}{2x-1}.

$$\frac{\left(2x-1\right)\left(2x+1\right)}{2x-1}$$

Cancel out 2x-1 in both numerator and denominator.

$$2x+1$$