Sum of the cubes of three successive natural number is divisible by

9

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27

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54

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99

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Solution

The correct option is B $9$
The three successive numbers are of the form $3 k, 3 k + 1, 3 k + 2$
Sum of their cubes $= (3 k)^{3} + (3 k + 1)^{3} + (3 k + 2)^{3}$
$= 27 k^{3} + 27 k^{3} + 1 + 9 k (3 k + 1) + 27 k^{3} + 8 + 18 k (3 k + 2)$
$= 81 k^{3} + 81 k^{2} + 45 k + 9$
$= 9 (9 k^{3} + 9 k^{2} + 5 k + 1) = 9 m$
Hence it is clearly divisible by 9.

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Similar questions

Prove that the sum of cubes of three consecutive natural numbers is always divisible by 3.

P (n)

n

\forall n \in N

n

P (n)

9

60

m

n + 1, n + 2, \dots, n + m

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