Question

[x²y²/a³b³]n

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Answer 1

Answer:

Hi ,

p = a²b³

q = a³b

HCF ( p,q ) = a²b

 [ ∵Product of the smallest power of each

     common prime factors in the numbers ]

LCM ( p , q ) = a³b³

[ ∵ Product of the greatest power of each 

  prime factors , in the numbers ]

Now ,

HCF ( p , q ) × LCM ( p , q ) = a²b × a³b³ 

= a∧5b∧4 --------( 1 )

[∵ a∧m × b∧n = a∧m+n ]

pq = a²b³ × a³b

= a∧5 b∧4 ---------------( 2 ) 

from ( 1 ) and ( 2 ) , we conclude 

HCF ( p , q ) × LCM ( p ,q ) = pq

Step-by-step explanation:

hope it helps

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