Question

Dont leave any links please

Answer 1

Answer:

m^2/n

Step-by-step explanation:

The following laws of exponent are useful to this problem:

$\displaystyle \large{ {(mn)}^{b} = {m}^{b} {n}^{b} } \\ \displaystyle \large{ {n}^{ - b} = \frac{1}{ {n}^{b} } }$

We are given the expression:

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } }$

Use the first law of exponent above.

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{(8)( \frac{1}{4}) } {n}^{ ( - 4)( \frac{1}{4}) } } \\ \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } {n}^{ - 1} }$

Make sure to recall the important necesscary fundamental math such as operation with negative numbers/integers, basic division, fraction, etc.

From the expression, apply the second law of exponent to n^-1.

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } ( \frac{1}{ {n}^{1} } )} \\ \displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = {m}^{2 } ( \frac{1}{ {n} } )}$

Multiply m in.

$\displaystyle \large{( {m}^{8} {n}^{ - 4} )^{ \frac{1}{4} } = \frac{ {m}^{2} }{ {n} } }$

Thus the answer is m^2/n.