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Find the cube root of each of the following natural numbers: (i) 343 (ii) 2744 (iii) 4913 (iv) 1728 (v) 35937 |
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Answer» (i) 343 By prime factorization method, = \(\sqrt[3]{343} \) = \(\sqrt[3]{7\times7\times7} = 7.\) (ii) 2744 By prime factorization method, = \(\sqrt[3]{2744}\) = \(\sqrt[3]{2\times2\times2\times7\times7\times7}\) = \(\sqrt[3]{2^3\times7^3} \) = \(2\times7\) = 14. (iii) 4913 By prime factorization method, = \(\sqrt[3]{1728}\) = \(\sqrt[3]{2\times2\times2\times2\times2\times2\times3\times3\times3}\) = \(\sqrt[3]{2^3\times2^3\times3^3}\) = (iv) 1728 By prime factorization method, = \(\sqrt[3]{1728}\) = \(\sqrt[3]{2\times2\times2\times2\times2\times2\times3\times3\times3}\) = \(\sqrt[3]{2^3{\times2^3}\times{3^3}}\) = \(2\times2\times3 = 12.\) (v) 35937 By prime factorization method, = \(\sqrt[3]{35937}\) = \(\sqrt[3]{3\times3\times3\times11\times11\times11}\) = \(\sqrt[3]{3^3\times11^3}\) = \(3\times11 = 33.\) |
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