Solve a^3-8/a^2-4+a/a^3+8

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

Evaluate

$$\frac{a^{4}+4a^{2}+a+16}{a^{3}+8}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{a^{3}-8}{a^{2}-4}.

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

Cancel out a-2 in both numerator and denominator.

$$\frac{a^{2}+2a+4}{a+2}+\frac{a}{a^{3}+8}$$

Factor a^{3}+8.

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

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

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

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

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

Do the multiplications in \left(a^{2}+2a+4\right)\left(a^{2}-2a+4\right)+a.

$$\frac{a^{4}-2a^{3}+4a^{2}+2a^{3}-4a^{2}+8a+4a^{2}-8a+16+a}{\left(a+2\right)\left(a^{2}-2a+4\right)}$$

Combine like terms in a^{4}-2a^{3}+4a^{2}+2a^{3}-4a^{2}+8a+4a^{2}-8a+16+a.

$$\frac{a^{4}+4a^{2}+a+16}{\left(a+2\right)\left(a^{2}-2a+4\right)}$$

Expand \left(a+2\right)\left(a^{2}-2a+4\right).

$$\frac{a^{4}+4a^{2}+a+16}{a^{3}+8}$$

Expand

$$\frac{a^{4}+4a^{2}+a+16}{a^{3}+8}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{a^{3}-8}{a^{2}-4}.

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

Cancel out a-2 in both numerator and denominator.

$$\frac{a^{2}+2a+4}{a+2}+\frac{a}{a^{3}+8}$$

Factor a^{3}+8.

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

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

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

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

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

Do the multiplications in \left(a^{2}+2a+4\right)\left(a^{2}-2a+4\right)+a.

$$\frac{a^{4}-2a^{3}+4a^{2}+2a^{3}-4a^{2}+8a+4a^{2}-8a+16+a}{\left(a+2\right)\left(a^{2}-2a+4\right)}$$

Combine like terms in a^{4}-2a^{3}+4a^{2}+2a^{3}-4a^{2}+8a+4a^{2}-8a+16+a.

$$\frac{a^{4}+4a^{2}+a+16}{\left(a+2\right)\left(a^{2}-2a+4\right)}$$

Expand \left(a+2\right)\left(a^{2}-2a+4\right).

$$\frac{a^{4}+4a^{2}+a+16}{a^{3}+8}$$