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However if one of the prime factors is a single Now, group factors in the triplets as 3 3 3´ ´ ´
4
4
factor or a double factor then the number 3 3 3 3 3 3 123
´
´
´
´
´
4
4
4
4
is not a perfect cube. 123 123
= 3 3 ´ 3 3 ´ 3 3
For example:
Since the prime factor 3 forms three
Examine if (i) 106480 and (ii) 19683 are perfect triplets
cubes.
so 19683 is a perfect cube.
(i) First we find the prime factors of the
number 106480 Example 3. What is the smallest number by
which 3087 may be multiplied so that the
2 106480
product is a perfect cube?
2 53240
Solution: Writing 3087 as a product of prime
2 26620
factors, we have:
2 13310
5 6655 3 3087
11 1351 3 1029
11 121 7 343
11 11
7 49
1
7 7
Prime Factors of 106480 = 2 ´ 2 ´ 2 ´ 2
5 11 11 11 1
´
´
´
Now, group factors in the triplets as 3087 = 3 ´ 3 ´ 7 ´ 7 ´ 7
´
2 ´ 2 ´ 2 ´ 2 ´ 5 11 11 11
´
´
Hence, to make it a perfect cube, it must be
One prime factor 2 and the prime factor 5 multiplied by 3.
are not parts of a triplet so 106480 is not a
perfect Example 4. What is the smallest number by
(ii) First we find the prime factors of the which 392 may be divided so that the quotient
number 19683 is a perfect cube?
3 19683 Solution: Writing 392 as a product of prime
factors, we have:
3 6561
3 2187 2 392
3 729 2 196

3 243 2 98
3 81 7 49

3 27 7 7
3 8 1

3 3 392 = 2 ´ 2 ´ 2 ´ 7 ´ 7
1 Clearly, to make it a perfect cube, it must be
divided by (7 × 7), i.e., 49.
Prime Factors of 19683 = 3 ´ 3 ´ 3 ´
3 ´ 3 ´ 3 ´ 3 ´ 3 ´ 3

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Mathematics In Focus - 8