If [tex]8a^{3b}[/tex] = 1 and a > 0, find the value of [tex]a^{2b} - \frac{1}{a^{b} }[/tex]

Some factoring lets us write[tex]a^{2b}-\dfrac1{a^b}=a^{2b}-a^{-b}=a^{-b}(a^{3b}-1)[/tex]Then[tex]8a^{3b}=1\implies a^{3b}=\dfrac18[/tex][tex]a^{2b}-\dfrac1{a^b}=a^{-b}\left(\dfrac18-1\right)=-\dfrac78a^{-b}=-\dfrac7{8a^b}[/tex]Taking the cube root to solve for [tex]b[/tex], we find[tex]\sqrt[3]{a^{3b}}=\sqrt[3]{\dfrac18}\implies a^b=\dfrac12[/tex]so ultimately[tex]a^{2b}-\dfrac1{a^b}=-\dfrac7{8\cdot\frac12}=-\dfrac74[/tex]