Solve 1/d+8-1-d/d^2-8d+64-(d^2+7d-8)/d^3+512=

$$\frac { 1 } { d + 8 } - \frac { 1 - d } { d ^ { 2 } - 8 d + 64 } - \frac { ( d ^ { 2 } + 7 d - 8 ) } { d ^ { 3 } + 512 } =$$

Evaluate

$$\frac{1}{d+8}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{d^{2}+7d-8}{d^{3}+512}.

$$\frac{1}{d+8}-\frac{1-d}{d^{2}-8d+64}-\frac{\left(d-1\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Cancel out d+8 in both numerator and denominator.

$$\frac{1}{d+8}-\frac{1-d}{d^{2}-8d+64}-\frac{d-1}{d^{2}-8d+64}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of d+8 and d^{2}-8d+64 is \left(d+8\right)\left(d^{2}-8d+64\right). Multiply \frac{1}{d+8} times \frac{d^{2}-8d+64}{d^{2}-8d+64}. Multiply \frac{1-d}{d^{2}-8d+64} times \frac{d+8}{d+8}.

$$\frac{d^{2}-8d+64}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{\left(1-d\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{d-1}{d^{2}-8d+64}$$

Since \frac{d^{2}-8d+64}{\left(d+8\right)\left(d^{2}-8d+64\right)} and \frac{\left(1-d\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{d^{2}-8d+64-\left(1-d\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{d-1}{d^{2}-8d+64}$$

Do the multiplications in d^{2}-8d+64-\left(1-d\right)\left(d+8\right).

$$\frac{d^{2}-8d+64-d-8+d^{2}+8d}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{d-1}{d^{2}-8d+64}$$

Combine like terms in d^{2}-8d+64-d-8+d^{2}+8d.

$$\frac{2d^{2}-d+56}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{d-1}{d^{2}-8d+64}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(d+8\right)\left(d^{2}-8d+64\right) and d^{2}-8d+64 is \left(d+8\right)\left(d^{2}-8d+64\right). Multiply \frac{d-1}{d^{2}-8d+64} times \frac{d+8}{d+8}.

$$\frac{2d^{2}-d+56}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{\left(d-1\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Since \frac{2d^{2}-d+56}{\left(d+8\right)\left(d^{2}-8d+64\right)} and \frac{\left(d-1\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2d^{2}-d+56-\left(d-1\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Do the multiplications in 2d^{2}-d+56-\left(d-1\right)\left(d+8\right).

$$\frac{2d^{2}-d+56-d^{2}-8d+d+8}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Combine like terms in 2d^{2}-d+56-d^{2}-8d+d+8.

$$\frac{d^{2}-8d+64}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Cancel out d^{2}-8d+64 in both numerator and denominator.

$$\frac{1}{d+8}$$

Expand

$$\frac{1}{d+8}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{d^{2}+7d-8}{d^{3}+512}.

$$\frac{1}{d+8}-\frac{1-d}{d^{2}-8d+64}-\frac{\left(d-1\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Cancel out d+8 in both numerator and denominator.

$$\frac{1}{d+8}-\frac{1-d}{d^{2}-8d+64}-\frac{d-1}{d^{2}-8d+64}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of d+8 and d^{2}-8d+64 is \left(d+8\right)\left(d^{2}-8d+64\right). Multiply \frac{1}{d+8} times \frac{d^{2}-8d+64}{d^{2}-8d+64}. Multiply \frac{1-d}{d^{2}-8d+64} times \frac{d+8}{d+8}.

$$\frac{d^{2}-8d+64}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{\left(1-d\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{d-1}{d^{2}-8d+64}$$

Since \frac{d^{2}-8d+64}{\left(d+8\right)\left(d^{2}-8d+64\right)} and \frac{\left(1-d\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{d^{2}-8d+64-\left(1-d\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{d-1}{d^{2}-8d+64}$$

Do the multiplications in d^{2}-8d+64-\left(1-d\right)\left(d+8\right).

$$\frac{d^{2}-8d+64-d-8+d^{2}+8d}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{d-1}{d^{2}-8d+64}$$

Combine like terms in d^{2}-8d+64-d-8+d^{2}+8d.

$$\frac{2d^{2}-d+56}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{d-1}{d^{2}-8d+64}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(d+8\right)\left(d^{2}-8d+64\right) and d^{2}-8d+64 is \left(d+8\right)\left(d^{2}-8d+64\right). Multiply \frac{d-1}{d^{2}-8d+64} times \frac{d+8}{d+8}.

$$\frac{2d^{2}-d+56}{\left(d+8\right)\left(d^{2}-8d+64\right)}-\frac{\left(d-1\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Since \frac{2d^{2}-d+56}{\left(d+8\right)\left(d^{2}-8d+64\right)} and \frac{\left(d-1\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2d^{2}-d+56-\left(d-1\right)\left(d+8\right)}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Do the multiplications in 2d^{2}-d+56-\left(d-1\right)\left(d+8\right).

$$\frac{2d^{2}-d+56-d^{2}-8d+d+8}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Combine like terms in 2d^{2}-d+56-d^{2}-8d+d+8.

$$\frac{d^{2}-8d+64}{\left(d+8\right)\left(d^{2}-8d+64\right)}$$

Cancel out d^{2}-8d+64 in both numerator and denominator.

$$\frac{1}{d+8}$$