How do you simplify #( -z^3 z^-6)/(-3z^2 )#?

1 Answer
EZ as pi
May 11, 2016

#1/(3z^5)#

Explanation:

There are several laws of indices in this expression, it does not matter which you regard as being the most important. You can apply any of the laws in any order.

Note: a negative sign in front is not the same as a negative sign in the index #rArr# treat them differently.

"add and subtract indices" #rArr (+z^(3-6-2))/3#

#= z^-5/3# = #" "1/(3z^5)#

Or: "a negative index indicates a reciprocal"

#+(z^3)/(3z^2z^6) = 1/(3z^(8-3) # = #" "1/(3z^5) #

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