Solve for: 3(x-4) <=-6

Expression: $$\frac { 9 } { a ^ { 3 } b } - \frac { a ^ { 2 } } { a ^ { 3 } b ^ { 3 } }$$

Cancel out $a^{2}$ in both numerator and denominator.

$$\frac{9}{a^{3}b}-\frac{1}{ab^{3}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $a^{3}b$ and $ab^{3}$ is $a^{3}b^{3}$. Multiply $\frac{9}{a^{3}b}$ times $\frac{b^{2}}{b^{2}}$. Multiply $\frac{1}{ab^{3}}$ times $\frac{a^{2}}{a^{2}}$.

$$\frac{9b^{2}}{a^{3}b^{3}}-\frac{a^{2}}{a^{3}b^{3}}$$

Since $\frac{9b^{2}}{a^{3}b^{3}}$ and $\frac{a^{2}}{a^{3}b^{3}}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{9b^{2}-a^{2}}{a^{3}b^{3}}$$