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Question
Which of the following numbers is a perfect cube?
Options
243
216
392
8640
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Solution
216
Explanation:
a. We have, 243
Resolving 243 into prime factors, we have
243 = 3 × 3 × 3 × 3 × 3
Grouping the factors in triplets of equal factors, we get
243 = (3 × 3 × 3) × 3 × 3
Clearly, in grouping, the factors in triplets of equal factors, we are left with two factors 3 × 3.
Therefore, 243 is not a perfect cube.
b. We have, 216
Resolving 216 into prime factors, we have
216 = 2 × 2 × 2 × 3 × 3 × 3
Grouping the factors in triplets of equal factors, we get
216 = (2 × 2 × 2) × (3 × 3 × 3)
Clearly, in grouping, the factors of triplets of equal factors, no factor is left over.
So, 216 is a perfect cube.
c. We have, 392
Resolving 392 into prime factors, we get
392 = 2 × 2 × 2 × 7 × 7
Grouping the factors in triplets of equal factors, we get
392 = (2 × 2 × 2) × 7 × 7
Clearly, in grouping, the factors in triplets of equal factors, we are left with two factors 7 × 7.
Therefore, 392 is not a perfect cube.
d. We have, 8640
Resolving 8640 into prime factors, we get
8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5
Grouping the factors in triplets of equal factors, we get
8640 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 5
Clearly, in grouping, the factors in triplets of equal factors, we are left with one factor 5.
Therefore, 8640 in not a perfect cube.
After solving, it is clear that 216 is correct.
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