find the smallest number by which 8788 must be multiplied to obtain a perfect cube
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Answered by
9
For a given number x. x cube would be a x×x×x which means x^3
A given natural number is a perfect cube if it can be expressed as a product of the triplets of equal factor.
hence write the given factor as a product of prime
8788= 2×2×(13×13×13)
in order to make it a perfect cube we have to multiply it with 2
(while multiplying we have to multiply with both the sides)
so
2×8788=2×2×2×13×13×13
17576=(2×13)^3
so the perfect cube is 26 and the smallest number by which 8788 must be multiplied to obtain a perfect cube is 2 .
Hope this helps you
A given natural number is a perfect cube if it can be expressed as a product of the triplets of equal factor.
hence write the given factor as a product of prime
8788= 2×2×(13×13×13)
in order to make it a perfect cube we have to multiply it with 2
(while multiplying we have to multiply with both the sides)
so
2×8788=2×2×2×13×13×13
17576=(2×13)^3
so the perfect cube is 26 and the smallest number by which 8788 must be multiplied to obtain a perfect cube is 2 .
Hope this helps you
Answered by
5
hey mate !
Here is your answer »»»»»»»»»»
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••••••••••••••••••••••••••••••••••••••••••••••••••
Step 1— Express 8788 as a product of its prime factors
8788 = 2 × 2 ×[13 × 13 × 13]
Step 2— Since "2" is the only group which is not in triplets, we need to multiply "2" to the number
8788 × 2 = 17576 which is a perfect cube
Therefore, the smallest number required to multiply to 8788 is 2.
••••••••••••••••••••••••••••••••••••••••••••••••••
Hope this helps
@PoojaBBSR
Here is your answer »»»»»»»»»»
↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
••••••••••••••••••••••••••••••••••••••••••••••••••
Step 1— Express 8788 as a product of its prime factors
8788 = 2 × 2 ×[13 × 13 × 13]
Step 2— Since "2" is the only group which is not in triplets, we need to multiply "2" to the number
8788 × 2 = 17576 which is a perfect cube
Therefore, the smallest number required to multiply to 8788 is 2.
••••••••••••••••••••••••••••••••••••••••••••••••••
Hope this helps
@PoojaBBSR
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