# The sum of cubes of three consecutive natural number is divisible by

A. 9

B. 27

C. 54

D. 99

Last updated date: 18th Mar 2023

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Answer

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Hint: Consider any three natural numbers and from this first find out the sum of the three consecutive natural numbers and then make use of the divisibility test and find out the answer.

Complete step-by-step answer:

Let us consider the three successive natural numbers to be a-1, a, a+1.

Now, we have to find out the cubes of these three consecutive numbers is divisible by what

So, the cubes of these three consecutive numbers would be \[{(a - 1)^3},{a^3},{(a + 1)^3}\]

Now, the sum of the cubes of these numbers would be

${(a - 1)^3} + {a^3} + {(a + 1)^3}$

Now , we know the formula which says ${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$

and ${\left( {a - b} \right)^3} = {a^3} + {b^3} - 3ab(a - b)$

So, making use of this formula , we can write

${(a - 1)^3} + {a^3} + {(a + 1)^3}$=${a^3} - 3{a^2} + 3a - 1 + {a^3} + {a^3} + 3{a^2} + 3a + 1$

= $3{a^3} + 6a$

=$3a({a^2} + 2)$

So, the sum of the cubes of three successive natural numbers

=$3a({a^2} + 2)$

In this equation either a or ${a^2} + 2$ has to be a multiple of 3

Now, from this we can clearly say that if a is multiple of 3 , then 3a is a multiple of 9

Else, if a is not a multiple of 3, then ${a^2} + 2$ is a multiple of 3

So, on combining these two results, we can say that $3a({a^2} + 2)$ is a multiple of 9 for all a$ \in $ N

So, from this we can say that the sum of cubes of 3 consecutive natural numbers is divisible by 9.

Note: The three consecutive natural numbers need not be a-1, a, a+1 only. We can take any other three consecutive natural numbers and solve it.

Complete step-by-step answer:

Let us consider the three successive natural numbers to be a-1, a, a+1.

Now, we have to find out the cubes of these three consecutive numbers is divisible by what

So, the cubes of these three consecutive numbers would be \[{(a - 1)^3},{a^3},{(a + 1)^3}\]

Now, the sum of the cubes of these numbers would be

${(a - 1)^3} + {a^3} + {(a + 1)^3}$

Now , we know the formula which says ${(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)$

and ${\left( {a - b} \right)^3} = {a^3} + {b^3} - 3ab(a - b)$

So, making use of this formula , we can write

${(a - 1)^3} + {a^3} + {(a + 1)^3}$=${a^3} - 3{a^2} + 3a - 1 + {a^3} + {a^3} + 3{a^2} + 3a + 1$

= $3{a^3} + 6a$

=$3a({a^2} + 2)$

So, the sum of the cubes of three successive natural numbers

=$3a({a^2} + 2)$

In this equation either a or ${a^2} + 2$ has to be a multiple of 3

Now, from this we can clearly say that if a is multiple of 3 , then 3a is a multiple of 9

Else, if a is not a multiple of 3, then ${a^2} + 2$ is a multiple of 3

So, on combining these two results, we can say that $3a({a^2} + 2)$ is a multiple of 9 for all a$ \in $ N

So, from this we can say that the sum of cubes of 3 consecutive natural numbers is divisible by 9.

Note: The three consecutive natural numbers need not be a-1, a, a+1 only. We can take any other three consecutive natural numbers and solve it.

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