Which of the following numbers are cubes of negative integers
Question:

Which of the following numbers are cubes of negative integers

(i) −64

(ii) −1056

(iii) −2197

(iv) −2744

(v) −42875

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 64 into prime factors, we get:

$64=2 \times 2 \times 2 \times 2 \times 2 \times 2$

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

$64=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$

It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube. This implies that 64 is also a perfect cube.

Now, collect one factor from each triplet and multiply, we ge

$2 \times 2=4$

This implies that 64 is a cube of 4.

Thus, $-64$ is the cube of $-4$.

(ii)

On factorising 1056 into prime factors, we get:

$1056=2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 11$

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

$1056=\{2 \times 2 \times 2\} \times 2 \times 2 \times 3 \times 11$

It is evident that the prime factors of 1056 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 1056 is not a perfect cube. This implies that $-1056$ is not a perfect cube as well.

(iii)

On factorising 2197 into prime factors, we get:

$2197=13 \times 13 \times 13$

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

$2197=\{13 \times 13 \times 13\}$

It is evident that the prime factors of 2197 can be grouped into triples of equal factors and no factor is left over. Therefore, 2197 is a perfect cube. This implies that $-2197$ is also a perfect cube.

Now, collect one factor from each triplet and multiply, we get 13.

This implies that 2197 is a cube of 13.

Thus, $-2197$ is the cube of $-13$.

(iv)

On factorising 2744 into prime factors, we get:

$2744=2 \times 2 \times 2 \times 7 \times 7 \times 7$

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

$2744=\{2 \times 2 \times 2\} \times\{7 \times 7 \times 7\}$

It is evident that the prime factors of 2744 can be grouped into triples of equal factors and no factor is left over. Therefore, 2744 is a perfect cube. This implies that $-2744$ is also a perfect cube.

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

$2 \times 7=14$

This implies that 2744 is a cube of 14.

Thus, $-2744$ is the cube of $-14$.

(v)

On factorising 42875 into prime factors, we get:

$42875=5 \times 5 \times 5 \times 7 \times 7 \times 7$

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

It is evident that the prime factors of 42875 can be grouped into triples of equal factors and no factor is left over. Therefore, 42875 is a perfect cube. This implies that $-42875$ is also a perfect cube.

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

$5 \times 7=35$

This implies that 42875 is a cube of 35.

Thus, $-42875$ is the cube of $-35$.

Administrator

Leave a comment

Please enter comment.
Please enter your name.