• # question_answer 3)                 Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. (i) 81                                      (ii) 128                                   (iii) 135                                  (iv) 192 (v) 704

By prime factorisation, $81=\underline{3\times 3\times 3}\times 3$                    | grouping the factors in triplets In the above factorisation 3 remains after grouping the 3's in triplets. Therefore, 81 is not a perfect cube. If we divide the number by 3, then in the prime factorisation of the quotient, this 3 will not remain. In that case, $81\div 3=\underline{3\times 3\times 3}$ $={{3}^{3}},$ which is a perfect cube    | by laws of exponents Hence, the smallest whole number by which 81 must be divided to obtain a perfect cube is 3. The resulting perfect cube is $27(={{3}^{3}})$. (ii) 128  2 128 2 64 2 32 2 16 2 8 2 4 2 2 1
By prime factorisation, $128=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 2$       | grouping the factors in triplets In the above factorisation, 2 remains after grouping the 2's in triplets. Therefore, 128 is not a perfect cube. If we divide the number by 2, then in the prime factorisation of the Quotient, this 2 will not remain. In that case, $128\div 2=\underline{2\,\times 2\times 2}\times \underline{2\times 2\times 2}$ $={{2}^{3}}\times {{2}^{3}}$ I by laws of exponents $={{(2\times 2)}^{3}}$ | by laws of exponents $={{4}^{3}},$ which is a perfect cube. Hence, the smallest whole number by which 128 must be divided to obtain a perfect cube is 2. The resulting perfect cube is $64(={{4}^{3}})$.  (iii) 135  3 135 3 45 3 15 5 5 1
By prime factorisation, $135=\underline{3\times 3\times 3}\times 5$ | grouping the factors in triplets. The prime factor 5 does not appear in a group of three. So 135 is not a perfect cube. In the factorisation 5 appears only ones. If we divide 135 by 5, then the prime factorisation of the quotient will not contain 5. In that case, $135\div 5=\underline{3\times 3\times 3}$ $={{3}^{3}},$ which is a perfect cube. | by laws of exponents Hence, the smallest whole number by which 135 must be divided to obtain a perfect cube is 5. The resulting perfect cube is $27(={{3}^{3}})$. (iv) 192  2 192 2 96 2 48 2 24 2 12 2 6 3 3 1
By prime factorisation, $192=\underline{2\,\times 2\times 2}\times \underline{2\times 2\times 2}\times 3$ | grouping the factors in triplets The prime factor 3 does not appear in a group of three. So 192 is not a perfect cube. In the factorisation of 192, 3 appears only once. So if we divide the number by 3, then the prime factorisation of the quotient will not contain 3. In that case, $192\div 3=\underline{2\,\times 2\times 2}\times \underline{2\times 2\times 2}$ $={{2}^{3}}\times {{2}^{3}}$     | by laws of exponents $={{(2\times 2)}^{3}}$                 | by laws of exponents $={{4}^{3}},$ which is a perfect cube. Hence, the smallest whole number by which 192 must be divided to obtain a perfect cube is 3. The resulting perfect cube is $64=({{4}^{3}})$. (v) 704  2 704 2 352 2 176 2 88 2 44 2 22 11 11 1
By prime factorisation, $704=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 11$ | grouping the factors in triplets, The prime factor 11 does not appears a group of three. So, 704 is not a perfect cube. In the factorisation 11 appears only one time. So if we divide 704 by 11, then the prime factorisation of the quotient will not contain 11 In that case, $704\div 11=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}$ $={{2}^{3}}\times {{2}^{3}}$ | by laws of exponents $={{(2\,\times 2)}^{3}}$ | by laws of exponents $={{4}^{3}},$ which is a perfect cube. Hence, the smallest whole number by which 704 must be divided to obtain a perfect cube is 11. The resulting perfect cube is $64=({{4}^{3}})$.