Solve 1/8(x-1)-1/8(x+3)+1/2(x+3)^2

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

Evaluate

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

Short Solution Steps

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

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

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

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

Do the multiplications in x+3-\left(x-1\right).

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

Combine like terms in x+3-x+1.

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

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

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

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

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

Do the multiplications in 4\left(x+3\right)+4\left(x-1\right).

$$\frac{4x+12+4x-4}{8\left(x-1\right)\left(x+3\right)^{2}}$$

Combine like terms in 4x+12+4x-4.

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

Factor the expressions that are not already factored in \frac{8x+8}{8\left(x-1\right)\left(x+3\right)^{2}}.

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

Cancel out 8 in both numerator and denominator.

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

Expand \left(x-1\right)\left(x+3\right)^{2}.

$$\frac{x+1}{x^{3}+5x^{2}+3x-9}$$

Expand

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

Short Solution Steps

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

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

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

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

Do the multiplications in x+3-\left(x-1\right).

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

Combine like terms in x+3-x+1.

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

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

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

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

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

Do the multiplications in 4\left(x+3\right)+4\left(x-1\right).

$$\frac{4x+12+4x-4}{8\left(x-1\right)\left(x+3\right)^{2}}$$

Combine like terms in 4x+12+4x-4.

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

Factor the expressions that are not already factored in \frac{8x+8}{8\left(x-1\right)\left(x+3\right)^{2}}.

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

Cancel out 8 in both numerator and denominator.

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

Expand \left(x-1\right)\left(x+3\right)^{2}.

$$\frac{x+1}{x^{3}+5x^{2}+3x-9}$$