Question

Find the least number by which 23328 should be multiplied to make it a perfect cube and the cube root of the product

Answer 1

Solution

TO FIND

The least number by which 23328 should be divided to make it a perfect cube

The cube root of each perfect cube

CALCULATION

The given number is 23328

\sf{23328 = 2 \times2 \times 2 \: \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 }23328=2×2×2×2×2×3×3×3×3×3×3

\implies \sf{23328 = {2}^{5} \times {3}^{6} }⟹23328=2

5

×3

6

{ {(3}^{2}) }^{3} }⟹23328=2

3

×2

2

×(3

2

)

3

\implies \sf{23328 = {2}^{3} \times {2}^{2} \times { 9 }^{3} }⟹23328=2

3

×2

2

×9

3

Hence in order to make a perfect cube 23328 should be divided by 4

Then the reduced number is 5832

\sf{5832 = {2}^{3} \times {9}^{3} \: }5832=2

3

×9

3

So that cube root of 5832 = 2 × 9 = 18

RESULT

Hence 4 is the least number by which 23328 should be divided to make it a perfect cube

Also the cube root of each reduced number 5832 is 18

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Answer 2

Answer:

least number = 2. cube root = 36

if we do its prime factor then we make the 3no. group of all factors 2 is less in the factors then we multiply the number with 2 and then do prime factors and we get answer 36