How do you simplify #\frac { n ^ { 0} } { ( n m ^ { 3} ) ^ { 2} \cdot 2m ^ { 3} n ^ { - 1} }#?

1 Answer
Jan 28, 2018

#1/(2m^9n)#

Explanation:

#(n^0)/((nm^3)^2*2m^3n^-1)#

Know that #n^0=1# for #ninRR, n!=0#

#:.=1/((nm^3)^2*2m^3n^-1)#

Also know that, #(ab)^n=a^nb^n#

#:.=1/(n^2m^6*2m^3n^-1)#

Another rule is that, #a^n*a^m=a^(n+m)#

#:.=1/(2m^9n^1)#

Final rule is that #a^1=a, ainRR#

#=1/(2m^9n)#