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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 $a, a^3$ is always greater than $a^2$.
(vi)If \( a \) and \( b \) are integers such that \( a^{2}>b^{2} \), then \( a^{3}>b^{3} . \).
(vii) If \( a \) divides \( b \), then \( a^{3} \) divides \( b^{3} \).
(viii) If \( a^{2} \) ends in 9, then \( a^{3} \) ends in \( 7 \).
(ix) If \( a^{2} \) ends in 5 , then \( a^{3} \) ends in 25 .
(x) If \( a^{2} \) ends in an even number of zeros, then \( a^{3} \) ends in an odd number of zeros.
To find:
We need to check whether the given statements are true or false.
Solution:
(i) Prime factorisation of 392 is,
$392=2\times2\times2\times7\times7$
$=2^3\times7^2$
Grouping the factors in triplets of equal factors, we see that two factors $7 \times 7$ are left.
Therefore, 392 is not a perfect cube.
The given statement is false.
(ii) Prime factorisation of 8640 is,
$8640=2\times2\times2\times2\times2\times2\times3\times3\times3\times5$
$=2^3\times2^3\times3^3\times5$
Grouping the factors in triplets of equal factors, we see that 5 is left.
Therefore, 8640 is not a perfect cube.
(iii) For every zero at the end in a number which when cubed gives three zeros at the end.
The given statement is true.
(iv) $4^3=4\times4\times4$ $=64$
Therefore,
The given statement is false.
(v) For $\frac{1}{2}$
$(\frac{1}{2})^3=\frac{1}{8}$
$(\frac{1}{2})^2=\frac{1}{4}$
$\frac{1}{8}<\frac{1}{4}$
If $n$ is a proper fraction, then the given statement is not possible.
Therefore,
The given statement is false.
(vi) For example,
$(-4)^2 = 16$ and $(-3)^2=9$
$16 > 9$
$(-4)^3 =-64$ and $(-3)^3 = -27$ $-64 < -27$
Therefore,
The given statement is false.
(vii) Let $a$ divides $b$, this implies,
$\frac{b}{a}= k$
$b=ak$
$\frac{b^3}{a^3} = \frac{(ak)^3}{a^3}$
$= \frac{a^3k^3}{a^3}$
$= k^3$
This is true for each value of $b$ and $a$.
Therefore,
The given statement is true.
(viii) For $a = 7$,
$7^2 = 49$
$7^3 = 343$
Therefore,
The given statement is false.
(ix) For $a = 15$,
$(15)^2 = 225$
$(15)^3 = 3375$
Therefore,
The given statement is false.
(x) For $a = 100$,
$(100)^2 = 10000$
$(100)^3 = 1000000$
Therefore,
The given statement is false.
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