Show that the following integers are cubes of negative integers.

Solution:

In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer $m,-m^{3}$ is the cube of $-m$.

(i)
On factorising 5832 into prime factors, we get:

$5832=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

On grouping the factors in triples of equal factors, we get:

$5832=\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\} \times\{3 \times 3 \times 3\}$

It is evident that the prime factors of 5832 can be grouped into triples of equal factors and no factor is left over. Therefore, 5832 is a perfect cube. This implies that $-5832$ is also a perfect cube.

Now, collect one factor from each triplet and multiply, we get:

$2 \times 3 \times 3=18$

This implies that 5832 is a cube of 18.

Thus, $-5832$ is the cube of $-18$.

(ii)
On factorising 2744000 into prime factors, we get:

$2744000=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7$

On grouping the factors in triples of equal factors, we get:

$2744000=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{5 \times 5 \times 5\} \times\{7 \times 7 \times 7\}$

It is evident that the prime factors of 2744000 can be grouped into triples of equal factors and no factor is left over. Therefore, 2744000 is a perfect cube. This implies that $-2744000$ is also a perfect cube.

Now, collect one factor from each triplet and multiply, we get: 

$2 \times 2 \times 5 \times 7=140$

This implies that 2744000 is a cube of 140.

Thus, $-2744000$ is the cube of $-140$.