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What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes?
(i) 675
(ii) 1323
(iii) 2560
(iv) 7803
(v) 107811
(vi) 35721
To do:
We have to find the smallest number by which the given numbers must be multiplied so that the products are perfect cubes.
Solution:
(i) Prime factorisation of $675=3\times3\times3\times5\times5$
Grouping the factors in triplets of equal factors, we find that $5 \times 5$ is not a complete triplet.
Therefore, by multiplying $675$ by 5, we get,
$675\times5=3\times3\times3\times5\times5\times5$
Hence, the smallest number by which the given number must be multiplied so that the product is a perfect cube is 5.
(ii) Prime factorisation of $1323=3\times3\times3\times7\times7$
Grouping the factors in triplets of equal factors, we find that $7 \times 7$ is not a complete triplet.
Therefore, by multiplying $1323$ by 7, we get,
$1323\times7=3\times3\times3\times7\times7\times7$
Hence, the smallest number by which the given number must be multiplied so that the product is a perfect cube is 7.
(iii) Prime factorisation of $2560=2\times2\times2\times2\times2\times2\times2\times2\times2\times5$
Grouping the factors in triplets of equal factors, we find that $5$ is not a complete triplet.
Therefore, by multiplying $2560$ by $5\times5=25$, we get,
$2560\times5\times5=2\times2\times2\times2\times2\times2\times2\times2\times2\times5\times5\times5$
Hence, the smallest number by which the given number must be multiplied so that the product is a perfect cube is 25.
(iv) Prime factorisation of $7803=3\times3\times3\times17\times17$
Grouping the factors in triplets of equal factors, we find that $17\times17$ is not a complete triplet.
Therefore, by multiplying $7809$ by $17$, we get,
$7809\times17=3\times3\times3\times17\times17\times17$
Hence, the smallest number by which the given number must be multiplied so that the product is a perfect cube is 17.
(v) Prime factorisation of $107811=3\times3\times3\times3\times11\times11\times11$
Grouping the factors in triplets of equal factors, we find that $3$ is not a complete triplet.
Therefore, by multiplying $107811$ by $3\times3=9$, we get,
$107811\times3\times3=3\times3\times3\times3\times3\times3\times11\times11\times11$
Hence, the smallest number by which the given number must be multiplied so that the product is a perfect cube is 9.
(vi) Prime factorisation of $35721=3\times3\times3\times3\times3\times3\times7\times7$
Grouping the factors in triplets of equal factors, we find that $7\times7$ is not a complete triplet.
Therefore, by multiplying $35721$ by $7$, we get,
$35721\times7=3\times3\times3\times3\times3\times3\times7\times7\times7$
Hence, the smallest number by which the given number must be multiplied so that the product is a perfect cube is 7.
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