From the question, we are asked to write
[tex](8a^{-3})^{-\frac{2}{3}}[/tex]in its simplest form.
Recall from the law of Indices that;
[tex](ab)^c=a^cb^c[/tex][tex]\Rightarrow(8a^{-3})^{-\frac{2}{3}}=8^{-\frac{2}{3}}\times(a^{-3})^{-\frac{2}{3}}[/tex]Also, recall that;
[tex]a^{-m}=\frac{1}{a^m},\text{ }(a^m)^n=a^{m\times n}[/tex][tex]\Rightarrow8^{-\frac{2}{3}}\times(a^{-3})^{-\frac{2}{3}}=\frac{1}{8^{\frac{2}{3}}}\times a^{-3\times-\frac{2}{3}}=\frac{1}{(\sqrt[3]{8})^2}\times a^2=\frac{1}{4}\times a^2=\frac{a^2}{4}[/tex]In the simplest form, the answer is (a^2)/4.
