Which of the following are perfect cubes?

Solution:

(i)
On factorising 64 into prime factors, we get

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

Group the factors in triples of equal factors as:

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

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.

(ii)
On factorising 216 into prime factors, we get:

$216=2 \times 2 \times 2 \times 3 \times 3 \times 3$

Group the factors in triples of equal factors as:

$216=\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\}$

It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.

(iii)
On factorising 243 into prime factors, we get:

$243=3 \times 3 \times 3 \times 3 \times 3$

Group the factors in triples of equal factors as:

$243=\{3 \times 3 \times 3\} \times 3 \times 3$

It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube

(iv)
On factorising 1000 into prime factors, we get:

$1000=2 \times 2 \times 2 \times 5 \times 5 \times 5$

Group the factors in triples of equal factors as:

$1000=\{2 \times 2 \times 2\} \times\{5 \times 5 \times 5\}$

It is evident that the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over. Therefore, 1000 is a perfect cube.

(v)
On factorising 1728 into prime factors, we get:

$1728=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

Group the factors in triples of equal factors as:

$1728=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\}$

It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.

(vi)
On factorising 3087 into prime factors, we get:

$3087=3 \times 3 \times 7 \times 7 \times 7$

Group the factors in triples of equal factors as:

$3087=3 \times 3 \times\{7 \times 7 \times 7\}$

It is evident that the prime factors of 3087 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube

(vii)
On factorising 4608 into prime factors, we get:

$4608=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3$

Group the factors in triples of equal factors as:

$4608=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times 3 \times 3$

It is evident that the prime factors of 4608 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 4608 is a not perfect cube.

(viii)
On factorising 106480 into prime factors, we get:

$106480=2 \times 2 \times 2 \times 2 \times 5 \times 11 \times 11 \times 11$

Group the factors in triples of equal factors as:

$106480=\{2 \times 2 \times 2\} \times 2 \times 5 \times\{11 \times 11 \times 11\}$

It is evident that the prime factors of 106480 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 106480 is a not perfect cube.

(ix)
On factorising 166375 into prime factors, we get:

$166375=5 \times 5 \times 5 \times 11 \times 11 \times 11$

Group the factors in triples of equal factors as:

$166375=\{5 \times 5 \times 5\} \times\{11 \times 11 \times 11\}$

It is evident that the prime factors of 166375 can be grouped into triples of equal factors and no factor is left over. Therefore, 166375 is a perfect cube.

(x)
On factorising 456533 into prime factors, we get:

$456533=7 \times 7 \times 7 \times 11 \times 11 \times 11$

Group the factors in triples of equal factors as:

$456533=\{7 \times 7 \times 7\} \times\{11 \times 11 \times 11\}$

It is evident that the prime factors of 456533 can be grouped into triples of equal factors and no factor is left over. Therefore, 456533 is a perfect cube.