Question 127
If'n' be any natural number. then by which largest number $$(n^3 - n)$$ is always divisible ?
Solution
$$(n^{3} - n) $$ can be written as n(n-1)(n+1)
for n to be any natural number, $$n^{3} - n$$ is a product 3 consecutive numbers starting from 1.
hence for any value of a min. product of 6 will be there hence it is always be divisible by 6.
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