Let $a=\frac{3}{5}$ and let $b=\frac{2}{3}$. Compute $a^{2}b^{-3}.$
To compute $a^{2}b^{-3}$, I need to recall the rules of exponents for powers of fractions. I remember that $(\frac{x}{y})^{n}=\frac{x^{n}}{y^{n}}$, so I can apply this rule to both $a^{2}$ and $b^{-3}$. This gives me $a^{2}b^{-3}=(\frac{3}{5})^{2}(\frac{2}{3})^{-3}=\frac{3^{2}}{5^{2}}\frac{2^{-3}}{3^{-3}}$. Now I need to simplify the fractions by multiplying the numerators and the denominators separately. This gives me $\frac{3^{2}}{5^{2}}\frac{2^{-3}}{3^{-3}}=\frac{3^{2}\cdot 2^{-3}}{5^{2}\cdot 3^{-3}}$. To simplify further, I need to recall the rules of exponents for negative powers. I remember that $x^{-n}=\frac{1}{x^{n}}$, so I can apply this rule to both $2^{-3}$ and $3^{-3}$. This gives me $\frac{3^{2}\cdot \frac{1}{2^{3}}}{5^{2}\cdot \frac{1}{3^{3}}}=\frac{3^{2}}{2^{3}}\frac{3^{3}}{5^{2}}$. To simplify further, I need to simplify the powers by multiplying the bases. This gives me $\frac{3^{2}}{2^{3}}\frac{3^{3}}{5^{2}}=\frac{9}{8}\frac{27}{25}$. Finally, I need to multiply the fractions by multiplying the numerators and the denominators separately. This gives me $\frac{9}{8}\frac{27}{25}=\frac{9\cdot 27}{8\cdot 25}=\frac{243}{200}$. 