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## Answers

so the numbers are 2a , 3a and 4a.

it is given that the sum of the cubes of 2a , 3a and 4a is 33957

so 99 times the cube of 'a' is 33957

__⇒cube of a is 343__

a = 7

so the numbers are 14 , 21 , 28.

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Let the numbers be 2x , 3x and 4x

(2x)³+(3x³)+(4x)³ = 33957

8x³+27x³+64x³= 33957

x³= 33957/99

x³=343

x= 7

so the numbers are 14,21,28

(2x)³+(3x³)+(4x)³ = 33957

8x³+27x³+64x³= 33957

x³= 33957/99

x³=343

x= 7

so the numbers are 14,21,28