prove that any number of form n^3-n is divisible by 6...please answer fast
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the value of n is equal to all even numbers put up the value is also divisible by 6 hope you like my question four examples in picture
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The product of n number of consecutive positive integers is always divisible by n. eg - 1×2×3×4 is divisible by 4 and 1×2×3×4×5×6 is divisible by 6.
∴n³−n = n(n²−1) = n(n+1)(n−1) = (n−1)n(n+1)
The above number: (n−1)n(n+1) is the product of three consecutive positive integers (n≥2) which is divisible by 3! =6
Hence, the number: n³−n is divisible by 6 for all positive integers n.
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∴n³−n = n(n²−1) = n(n+1)(n−1) = (n−1)n(n+1)
The above number: (n−1)n(n+1) is the product of three consecutive positive integers (n≥2) which is divisible by 3! =6
Hence, the number: n³−n is divisible by 6 for all positive integers n.
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