Please answer all 7 questions

$$\begin{gathered} Dear Student \\ Q3) \\ \frac{9^{ {\displaystyle \frac{1}{ 3 }}} \times 27 { }^{-12} }{3^{ {\displaystyle \frac{1}{ 6 }}} \times 3^{ {\displaystyle \frac{-2}{3}}}} = \frac{{\left({\left(3\right)}^2\right)}^{{\displaystyle \frac{1}{3}}} \times {\left(3^3\right)}^{-12}}{3^{{\displaystyle \frac{1}{6 }} }\times 3^{{\displaystyle \frac{-2}{3}}}} (u\sin g 9 = 3^2 and 27 =3^3 ) \\ = \frac{3^{ {\displaystyle \frac{2}{3}} }\times 3^{-36}}{3\left({}^{{\displaystyle \frac{1}{6}}+{\displaystyle \frac{-2}{3}}}\right)} (u\sin g {\left(a^n\right)}^m = a^{nm} , b^n \times b^m =b^{n+m} ) \\ = \frac{3^{{\displaystyle \frac{2}{3}}+\left(-36\right)}}{3^{{\displaystyle \frac{1-4}{6}}}} = \frac{3^{{\displaystyle \frac{2-108}{3}}}}{3^{{\displaystyle \frac{-3}{6}}}} = \frac{3^{{\displaystyle \frac{-106}{3}}}}{3^{{\displaystyle \frac{-1}{2}}}} \\ = 3^{\frac{-106}{3} -\left(\frac{-1}{2}\right)} (u\sin g \frac{a^n}{a^m} = a^{n-m}) \\ =3^{\frac{-106}{3}+\frac{1}{2}} \\ =3^{\frac{-212+3}{6}} = 3^{-\frac{209}{6}} \end{gathered}$$

Kindly ask the remaining question in a separate thread


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