Question

The cube root of the sum of three consecutive odd natural number is equal to the square root of cube root
of 225. Find the sum of cubes of the number. [

Answer 1

Answer:

n

3

+(n+1)

3

+(n−1)

3

=3(n)

3

+6n

=3n(n

2

+2)

=3n(n

2

−1+3)

=3(n(n−1)(n+1)+3n)

n(n−1)(n+1) and 3n are both divisible by 3. Hence their sum is also divisible by 3.

Therefore the whole number is divisible by 9.

Answer 2

Answer:

495

Step-by-step explanation:

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