How do you multiply #2x ^ { 2} y ^ { 5} ( 9x ^ { 3} y ) ^ { - 3}#?

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Answer 1 · 2017-08-22T01:00:22

See below.

Let's deal with the value inside the parenthesis first.

#2x ^ { 2} y ^ { 5} ( 9x ^ { 3} y ) ^ { - 3}#

Remember that #x^-a=1/x^a#

#( 9x ^ { 3} y ) ^ { - 3}=9^-3x^-9y^-3=1/(729x^9y^3)#

Combining,

#2x ^ { 2} y ^ { 5} ( 9x ^ { 3} y ) ^ { - 3}=2x ^ { 2} y ^ { 5} *1/(729x^9y^3)#

#x^a/x^b=x^(a-b)#

So,

#2x ^ { 2} y ^ { 5} *1/(729x^9y^3)=\frac(2y^2)(729x^7)#

Since #2# and #9# have no common multiples, this is our solution.