#### Answer

$\frac{16b + 15}{24b^3}$

#### Work Step by Step

To solve: $\frac{4}{6b^2} + \frac{5}{8b^3}$, you will first need to make sure both fractions have common denominators. To do this, we must find the LCD, or the Least Common Multiple, by writing out each as products of prime factors.
$6b^2 = 2 \times 3 \times b \times b$
$8b^3 = 2 \times 2 \times 2 \times b \times b\times b$
The Least Common Multiple, or LCD, then is $2\times 2\times 2\times 3\times b\times b\times b = 24b^3$
Now\times we can rewrite the fractions using the new LCD we found.
$\frac{4}{6b^2} + \frac{5}{8b3}$ = $\frac{4\times 2\times 2\times b}{6b^2 \times 2 \times 2 \times b} + \frac{5 \times3}{8b^3 \times 3}$ =$\frac{16b}{24b^3} + \frac{15}{24b^3}$
Now that we have common denominators, we can add the numerators:
$\frac{16b}{24b^3} + \frac{15}{24b^3}$ = $\frac{16b + 15}{24b^3}$