Question

HCF of a³b +3a³b² and (ab)²-4a²b²​

Answer 1

Answer:

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Step-by-step explanation:

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

Answer 2

Step-by-step explanation:

a³b +3a³b² = a³b (1++3b)

-3a²b²

HCF = ab

HCF of a³b +3a³b² and (ab)²-4a²b²= ab