Solve b-3/8b^4-3b/b^2-9

$$\frac { b - 3 } { 8 b ^ { 4 } } - \frac { 3 b } { b ^ { 2 } - 9 }$$

Evaluate

$$\frac{27-9b-3b^{2}+b^{3}-24b^{5}}{8\left(b^{2}-9\right)b^{4}}$$

Short Solution Steps

Factor b^{2}-9.

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

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

$$\frac{\left(b-3\right)\left(b-3\right)\left(b+3\right)}{8\left(b-3\right)\left(b+3\right)b^{4}}-\frac{3b\times 8b^{4}}{8\left(b-3\right)\left(b+3\right)b^{4}}$$

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

$$\frac{\left(b-3\right)\left(b-3\right)\left(b+3\right)-3b\times 8b^{4}}{8\left(b-3\right)\left(b+3\right)b^{4}}$$

Do the multiplications in \left(b-3\right)\left(b-3\right)\left(b+3\right)-3b\times 8b^{4}.

$$\frac{b^{3}-9b-3b^{2}+27-24b^{5}}{8\left(b-3\right)\left(b+3\right)b^{4}}$$

Expand 8\left(b-3\right)\left(b+3\right)b^{4}.

$$\frac{b^{3}-9b-3b^{2}+27-24b^{5}}{8b^{6}-72b^{4}}$$

Expand

$$-\frac{24b^{5}-b^{3}+3b^{2}+9b-27}{8\left(b^{2}-9\right)b^{4}}$$

Short Solution Steps

Factor b^{2}-9.

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

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

$$\frac{\left(b-3\right)\left(b-3\right)\left(b+3\right)}{8\left(b-3\right)\left(b+3\right)b^{4}}-\frac{3b\times 8b^{4}}{8\left(b-3\right)\left(b+3\right)b^{4}}$$

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

$$\frac{\left(b-3\right)\left(b-3\right)\left(b+3\right)-3b\times 8b^{4}}{8\left(b-3\right)\left(b+3\right)b^{4}}$$

Do the multiplications in \left(b-3\right)\left(b-3\right)\left(b+3\right)-3b\times 8b^{4}.

$$\frac{b^{3}-9b-3b^{2}+27-24b^{5}}{8\left(b-3\right)\left(b+3\right)b^{4}}$$

Expand 8\left(b-3\right)\left(b+3\right)b^{4}.

$$\frac{b^{3}-9b-3b^{2}+27-24b^{5}}{8b^{6}-72b^{4}}$$