If HCF and LCM of two numbers are respectively (n-1) and (n^2-1) (n^2-4), then the product of two numbers will be-:
Please upload correct answer with explanation.
Answers
Given : HCF of two numbers = n - 1
LCM = (n² - 1)(n² - 4)
To Find : product of two numbers
Solution:
Two number a , b
HCF (a , b) * LCM (a , b) = product of two numbers
HCF (a , b) = n - 1
LCM (a , b) = (n² - 1)(n² - 4)
product of two numbers = (n -1 ) (n² - 1)(n² - 4)
using x² - y² = (x + y)(x - y)
= (n -1 )(n+1 )(n - 1)(n² - 4)
= (n² - 4) (n + 1)(n - 1)²
Hence Option C is correct
product of two numbers is (n² - 4) (n + 1)(n - 1)²
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Answer:
Given : HCF of two numbers = n - 1
LCM = (n² - 1)(n² - 4)
To Find : product of two numbers
Solution:
Two number a , b
HCF (a , b) * LCM (a , b) = product of two numbers
HCF (a , b) = n - 1
LCM (a , b) = (n² - 1)(n² - 4)
product of two numbers = (n -1 ) (n² - 1)(n² - 4)
using x² - y² = (x + y)(x - y)
= (n -1 )(n+1 )(n - 1)(n² - 4)
= (n² - 4) (n + 1)(n - 1)²
Hence Option C is correct
product of two numbers is (n² - 4) (n + 1)(n - 1)²
Learn More:
Find LCM and HCF of the following pairs of integers and verify that ...