Write true (T) or false (F) for the following statements:

Solution:

(i) False

On factorising 392 into prime factors, we get:

$392=2 \times 2 \times 2 \times 7 \times 7$

 

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

$392=\{2 \times 2 \times 2\} \times 7 \times 7$

It is evident that the prime factors of 392 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 392 is not a perfect cube.

(ii) True

On factorising 8640 into prime factors, we get:

$8640=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$

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

$8640=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\} \times 5$

It is evident that the prime factors of 8640 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 8640 is not a perfect cube.

(iii) True

Because a perfect cube always ends with multiples of 3 zeros, e.g., 3 zeros, 6 zeros etc.

(iv) False.

64 is a perfect cube, and it ends with 4.

(v) False

It is not true for a negative integer. Example: $(-5)^{2}=25 ;(-5)^{3}=-125 \Rightarrow(-5)^{3}<(-5)^{2}$

(vi) False

It is not true for negative integers. Example: $(-5)^{2}>(-4)^{2}$ but $(-5)^{3}<(-4)^{3}$

(vii) True

$\because$ a divides $b$

$\therefore \frac{b^{3}}{a^{3}}=\frac{b \times b \times b}{a \times a \times a}=\frac{(a k) \times(a k) \times(a k)}{a \times a \times a}$

$\because$ a divides $b$

$\therefore b=a k$ for some $k$

$\therefore \frac{b^{3}}{a^{3}}=\frac{(a k) \times(a k) \times(a k)}{a \times a \times a}=k^{3} \Rightarrow b^{3}=a^{3}\left(k^{3}\right)$

$\therefore a^{3}$ divides $b^{3}$

(viii) Fals

a3 ends in 7 if a ends with 3.

But for every a2 ending in 9, it is not necessary that a is 3.

E.g., 49 is a square of 7 and cube of 7 is 343.

(ix) False

$\because 35^{2}=1225$ but $35^{3}=42875$

(x) False

$\because 100^{2}=10000$ and $100^{3}=100000$