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

Question:

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

(i) 392 is a perfect cube.

(ii) 8640 is not a perfect cube.

(iii) No cube can end with exactly two zeros.

(iv) There is no perfect cube which ends in 4.

(v) For an integer aa3 is always greater than a2.

(vi) If a and b are integers such that a2 > b2, then a3 > b3.

(vii) If a divides b, then a3 divides b3.

(viii) If a2 ends in 9, then a3 ends in 7.

(ix) If a2 ends in 5, then a3 ends in 25.

(x) If a2 ends in an even number of zeros, then a3 ends in an odd number of zeros.

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 $100^{3}=100000$