Math, asked by pawansingh1082, 4 hours ago

Write cube of 5 natural numbers which are multiples of 3 and verify the cube of natural number, which is a multiple of 3 is a multiple of 27.​

Answers

Answered by arushi2981
2

Answer:

(3)

3

=3×3×3=27

(6)

3

=6×6×6=216

(9)

3

=9×9×9=729

(12)

3

=12×12×12=1728

(15)

3

=15×15×15=3375

Verification:

(3)

3

=27=27×1

(6)

3

=216=27×8

(9)

3

=729=27×27

(12)

3

=1729=27×64

(15)

3

=3375=27×125

∴ 'The cube of natural number, which is a multiple of 3 is a multiple of 27'.

Answered by OoINTROVERToO
10

Step-by-step explanation:

Let say 5 multiples of 3 = 6, 9, 12, 15, 21

Cube of multiple of 3 are as follows:

  • 6³ = 216
  • 9³ = 729
  • 12³ = 1728
  • 15³ = 3375
  • 21³ = 9261

Now, we have to verify cube of multiple of 3 is also multiple of 3

So we check above mentioned cubes are multiple of 3 or not.

If those nos. divided by 3 then they are multiple of 3

  • 2163 = 72 gives 216 is multiples of 3
  • 7293 = 243 gives 729 is multiples of 3
  • 1728 + 3 = 576 gives 1728 is multiples of 3
  • 33753 = 1125 gives 3375 is multiples of 3
  • 92613 = 3087 gives 9261 is multiples of 3

__Hence Verified

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