Answer:
[tex] \frac{ {4a}^{4} }{b} [/tex]
solution,
[tex] \sqrt{16 {a}^{8} {b}^{ - 2} } \\ [/tex]
Use negative power rule:
[tex] {x}^{ - a} = \frac{1}{ {x}^{a} } [/tex]
[tex] \sqrt{ {16}^{8} \times \frac{1}{ {b}^{2} } } \\ [/tex]
Simplify:
[tex] \sqrt{ \frac{ {16a}^{8} }{ {b}^{2} } } \\ = \frac{ \sqrt{ {16a}^{8} } }{ \sqrt{ {b}^{2} } } \\ [/tex]
Use this rule:
[tex] \sqrt{ab} = \sqrt{a} . \sqrt{b} [/tex]
[tex] \frac{ \sqrt{16. \sqrt{ {a}^{8} } } }{ \sqrt{ {b}^{2} } } [/tex]
Since, 4*4=16 ,the square root of 16 is 4
[tex] \frac{ \sqrt{ {4}^{2} } \sqrt{ {a}^{8} } }{ \sqrt{ {b}^{2} } } \\ = \frac{4 \sqrt{ {a}^{8} } }{ \sqrt{ {b}^{2} } } [/tex]
Simplify:
[tex] \frac{4 \: \sqrt{ {(a}^{4) ^{2} } } }{ \sqrt{ {b}^{2} } } \\ = \frac{4 {a}^{4} }{b} [/tex]
Hope this helps...
Good luck on your assignment...
