Question

Find the least number by which 23328 should be divided to make it a perfect cube also find the cube root of each perfect cube​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

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

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

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

$\implies \sf{23328 = {2}^{3} \times {2}^{2} \times { {(3}^{2}) }^{3} }$

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

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

Then the reduced number is 5832

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

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:

this is the correct

Step-by-step explanation:

the above one