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Is 500 a perfect cube?


500 = 5 x 5 x 5 x 2 x 2

∵ In the above prime factorisation 2 x2 remain after grouping the prime factors in
 triples.


500 = 5 x 5 x 5 x 2 x 2∵ In the above prime factorisation 2 x2 rema

∴ 500 is not a perfect cube.

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Is 31944 a perfect cube? If not then by which smallest natural number should 31944 be divided so that the quotient is a perfect cube?

We hve 31944 = 2 x 2 x 2 x 3 x 11 x 11 x 11

Since, the prime factors of 31944 do not appear in triples as 3 is left over.


We hve 31944 = 2 x 2 x 2 x 3 x 11 x 11 x 11Since, the prime factors o
∴ 31944 is  not a perfect cube, Obviously, 31944 divided by  will be a perfect cube

i.e.          [31944] divided by 3 = [2 x 2 x 2 x 3 x 11 x 11 x 11]divided by 3

or 10648 = 2 x 2 x 2 x 11 x 11 x 11

∴ 10648 is a perfect cube.

Thus, the required  least number = 3.

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Is 216 a perfect cube? What is the number whose cube is 216?


We have 216

We have 216i.e. Resolving 216 into prime factors we have: 216 = 2 x 2

i.e. Resolving 216 into prime factors we have: 216 = 2 x 2 x 2 x 3x 3 x 3

i.e. 216 can be resolved in to such prime factor which can be grouped into triple of equal of factor and no factor is left over.

∴ 216 is a perfect cube

We have 216i.e. Resolving 216 into prime factors we have: 216 = 2 x 2
Thus, the required number is 6, whose cube is 216.

 
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Show that —1728 is a perfect cube. Also, find the number whose cube is – 1728. 


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Is 1372 a perfect cube? If not, find the smallest natural number by which 1372 must be multipled so that the product is a perfect cube.


We have 1372 = 2 x 2 x 7 x 7 x 7

Since, the prime factor 2 does not appear in a group of triples.  


We have 1372 = 2 x 2 x 7 x 7 x 7Since, the prime factor 2 does not ap
∴ 1372  is  not a  perfect cube.

Obviously, to make it a perfect cube we need one more 2 as its factor.

i.e.            [1372] x 2 = [2 x 2 x 7 x 7 x 7] x 2

or                2744 = 2 x 2 x 2 x 7 x 7 x 7

which is a perfect cube.

Thus, the required smallest number = 2.



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