Question

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

Answer 1

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.