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 a, a³ 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.

Answer 1

(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:
$392=\left\{2\times 2\times 2\right\}\times 7\times 7$
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:
$8640=\left\{2\times 2\times 2\right\}\times \left\{2\times 2\times 2\right\}\times \left\{3\times 3\times 3\right\}\times 5$
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: ${\left(-5\right)}^2=25;{\left(-5\right)}^3=-125\Rightarrow {\left(-5\right)}^3<{\left(-5\right)}^2$
(vi) False

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

(vii) True

$\because$ a divides b
$\therefore$ $\frac{b^3}{a^3}=\frac{b\times b\times b}{a\times a\times a}=\frac{\left(ak\right)\times \left(ak\right)\times \left(ak\right)}{a\times a\times a}$

$\because$ a divides b
$\therefore$ b = ak for some k

$\therefore$$\frac{b^3}{a^3}=\frac{\left(ak\right)\times \left(ak\right)\times \left(ak\right)}{a\times a\times a}=k^3\Rightarrow b^3=a^3\left(k^3\right)$
$\therefore$ a³ divides b³

(viii) False

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

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

(x) False

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