Question:
Using prime factorisatioji, find which of the following are perfect cubes,
(a) 128
(b) 343
(c) 729
(d) 1331
Solution:
(a) We have, 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Since, 2 remains after grouping in triplets.
So, 128 is not a perfect cube.
(b) We have, 343 = 7 x 7 x 7
Since, the prime factors appear in triplets.
So, 343 is a perfect cube.
(c) We have, 729 = 3 x 3 x 3 x 3 x 3 x3
Since, the prime factors appear in triplets.
So, 729 is a perfect cube.
(d) We have, 1331 =11x11x11
Since, the prime factors appear in triplets.
So, 1331 is a perfect cube.
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