Question

Find the smallest number by which 8640 must be divided so that the quotient is a perfect cube. also find the cube root of the number so obtained.

Answer 1

5 is the smallest number by which 8640 must be divided so that the quotient is a perfect cube.

8640=2.2.2.2.2.2.3.3.3.5

therfore smallest number=5

cube root obtained=8640/5=1728.

Answer 2

Answer:

The required smallest number which 8640 must be divided so that the quotient is a perfect cube is 5.

The required cube root of 1728 is 12.

Step-by-step explanation:

To find : The smallest number by which 8640 must be divided so that the quotient is a perfect cube. also find the cube root of the number so obtained.

Solution :

First we factor the number 8640,

$8640=2 \times 2\times2\times2\times2\times2 \times3\times3\times3 \times 5$

Making a pair of 3,

$8640=2^3\times2^3 \times3^3 \times 5$

As 5 left alone which means if we divide 8640 by 5 we the the number having a perfect cube.

So, The required smallest number which 8640 must be divided so that the quotient is a perfect cube is 5.

Now, Divide by 5

$\frac{8640}{5}=\frac{2^3\times2^3 \times3^3 \times 5}{5}$

$1728=(2\times2\times3)^3$

$1728=(12)^3$

Therefore, The required cube root of 1728 is 12.