How do you simplify #((4n)/(2n^2))^3#?

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Answer 1 · 2015-10-27T04:23:27

Here,

First you can open the bracket and multiply the power of n with 3 both in the numerator and denominator.

= #((4n)/(2n^2))^3#

= #((4^3n^3)/(2^3n^6))#

Now you can cancel the three n and 8 as it is present in both numerator and denominator.

= #((64*n*n*n )/(8*n*n*n*n*n*n))#

= #8/n^3#