• # question_answer 2)                 Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (i) 243                                    (ii) 256                                   (iii) 72                                    (iv) 675 (v) 100

By prime factorisation, $243=\underline{3\times 3\times 3}\times 3\times 3$ | grouping the factors in triplets The prime factor 3 does not appear in a group of three. Therefore, 243 is not a perfect cube. To make it a cube, we need one more 3. In that case $243\times 3=\underline{3\times 3\times 3}\times \underline{3\times 3\times 3}$ = 729, which is a perfect cube. Hence, the smallest number by which 243 should be multiplied to make a perfect cube is 3. (ii) 256  2 256 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1
By prime factorisation, $256=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 2\times 2$               | grouping the factors in triplets In the above factorisation 2 remains after grouping 2's in triplets. Therefore, 128 is not a perfect cube. To make it a perfect cube, we need one 2's more. In that case, $256\times 2=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}$ $={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}$                     | by laws of exponents $={{(2\times 2\times 2)}^{3}}$                                | by laws of exponents $={{8}^{3}}=512,$ which is a perfect cube. Hence, the smallest number by which 256 must be multiplied to obtain a perfect cube is 2. The resulting perfect cube is $512=({{8}^{3}})$. (iii) 72    2 72 2 36 2 18 3 9 3 3 1
By prime factorisation, $72=\underline{2\times 2\times 2}\times 3\times 3$ | grouping the factors in triplets The prime factors 3 does not appear in a group of three. Therefore, 72 is not a perfect cube. To make it a cube, we need one more 3. In that case, $72\times 3=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}$ $={{2}^{3}}\times {{3}^{3}}$     | by laws of exponents $={{(2\ \times 3)}^{3}}$              | by laws of exponents $={{6}^{3}},$ which is a perfect cube. Hence, the smallest number by which 72 must be multiplied to obtain a perfect cube is 3. (iv) 675    3 675 3 225 3 75 5 25 5 5 1
By prime factorisation, $675=\underline{3\times 3\times 3}\times 5\times 5$ | grouping the factors in triplets The prime factor 5 does not appear in a group of three. Therefore, 675 is not a perfect cube. To make it a cube, we need one more 5. In that case, $675\times 5=\underline{3\times 3\times 3}\times \underline{5\times 5\times 5}$ $={{3}^{3}}\times {{5}^{3}}$     | by laws of exponents $={{(3\times 5)}^{3}}$                 | by laws of exponents $={{15}^{3}},$ which is a perfect cub Hence, the smallest number by which 675 must be multiplied to obtain a perfect cube is 5. The resulting perfect cube is 3375 $(={{15}^{3}})$ (v) 100  2 100 2 50 5 25 5 5 1
By prime factorisation, $100=2\times 2\times 5\times 5$ | grouping the factors in triplets The prime factors 2 and 5 do not appear in a group of three. Therefore, 100 is not a perfect cube. To make it a perfect cube, we need one 2 and one 5 more. In that case, $100\,\times 2\times 5=\underline{2\times 2\times 2}\times \underline{5\times 5\times 5}$ $={{2}^{3}}\times {{5}^{3}}$     | by laws of exponents $={{(2\times 5)}^{3}}$                 | by laws of exponents $={{10}^{3}},$ which is a perfect cube. Hence, the smallest number by which 100 must be multiplied to obtain a perfect cube is $2\times 5=10$. The resulting perfect cube is $1000(={{10}^{3}})$.