False. Solution: 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.
True. Solution: $\because$ a divides b $\therefore\frac{\text{b}^3}{\text{a}^3}=\frac{\text{b}\times\text{b}\times\text{b}}{\text{a}\times\text{a}\times\text{a}}=\frac{\text{(ak)}\times\text{(ak)}\times\text{(ak)}}{\text{a}\times\text{a}\times\text{a}}$ $\because$ a divides b $\therefore$ b = ak fore some k $\therefore\frac{\text{b}^3}{\text{a}^3}=\frac{\text{(ak)}\times\text{(ak)}\times\text{(ak)}}{\text{a}\times\text{a}\times\text{a}}=\text{k}^3$ $\Rightarrow\text{k}^3=\text{b}^3=\text{a}^3(\text{k}^3)$ $\therefore$ a3 divides b3
True. Solution: On factorising 8640 into prime factors, we got: 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 On grouping the factors in triples of equal factors, we get: 8640 = {2 × 2 × 2} × {2 × 2 × 2} × {3 × 3 × 3} × 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.
False. Solution: On factorising 392 into prime factors, we got: 392 = 2 × 2 × 2 × 7 × 7 On grouping the factors in triples of equal factors, we get: 392 = {2 × 2 × 2} × 7 × 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.