Find the HCF of : 15 a3b2c3
, 12 a4b c4
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For H.C.F, we need to take the product of the smallest power of each common prime factors in the numbers. So,
15 a3b2c3 , 12 a4bc4
= a3bc3
Hence, the H.C.F is a3bc3
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Answer:
15a³b²c³ = 3*5*a*a*a*b*b*c*c*c
12 a4 bc4 = 2*2*3*a*a*a*a*b*c*c*c*c
H.C.F = 3a³bc³
200percent correct answer.
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