Solve 5/3b^3-2b^2-5-2/b^3-2?

$$\frac { 5 } { 3 b ^ { 3 } - 2 b ^ { 2 } - 5 } - \frac { 2 } { b ^ { 3 } - 2 } ?$$

Evaluate

$$\frac{\left(4-b\right)b^{2}}{\left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3b^{3}-2b^{2}-5 and b^{3}-2 is \left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right). Multiply \frac{5}{3b^{3}-2b^{2}-5} times \frac{b^{3}-2}{b^{3}-2}. Multiply \frac{2}{b^{3}-2} times \frac{3b^{3}-2b^{2}-5}{3b^{3}-2b^{2}-5}.

$$\frac{5\left(b^{3}-2\right)}{\left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right)}-\frac{2\left(3b^{3}-2b^{2}-5\right)}{\left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right)}$$

Since \frac{5\left(b^{3}-2\right)}{\left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right)} and \frac{2\left(3b^{3}-2b^{2}-5\right)}{\left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5\left(b^{3}-2\right)-2\left(3b^{3}-2b^{2}-5\right)}{\left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right)}$$

Do the multiplications in 5\left(b^{3}-2\right)-2\left(3b^{3}-2b^{2}-5\right).

$$\frac{5b^{3}-10-6b^{3}+4b^{2}+10}{\left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right)}$$

Combine like terms in 5b^{3}-10-6b^{3}+4b^{2}+10.

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

Expand \left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right).

$$\frac{-b^{3}+4b^{2}}{3b^{6}-2b^{5}-11b^{3}+4b^{2}+10}$$

Differentiate w.r.t. b

$$\frac{b\left(9b^{7}-52b^{6}+24b^{5}+40b^{3}-30b+80\right)}{\left(\left(b^{3}-2\right)\left(3b^{3}-2b^{2}-5\right)\right)^{2}}$$