1.Find the cube root of each of the following numbers by prime factorisation method.
(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125


Answer :

(i) We have the prime factors of 64 as:

\(\qquad \begin{array}{c|lcr} 2 & 64\\ \hline 2 & 32\\ \hline 2 & 16\\ \hline 2 & 8\\ \hline 2 & 4\\ \hline 2 & 2\\ \hline & 1\\ \end{array} \)

\(256 =2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^3 \times 2^3\)

∴\(\sqrt[3]{64}=2 \times 2=4\)

Hence, the cube root of 64 is 4.

(ii) We have the prime factors of 512 as:

\(\qquad \begin{array}{c|lcr} 2 & 512\\ \hline 2 & 256\\ \hline 2 & 128\\ \hline 2 & 64\\ \hline 2 & 32\\ \hline 2 & 16\\ \hline 2 & 8\\ \hline 2 & 4\\ \hline 2 & 2\\ \hline & 1\\ \end{array} \)

\(256 =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^3 \times 2^3 \times 2^3\)

∴\(\sqrt[3]{512}=2 \times 2 \times 2=8\)

Hence, the cube root of 512 is 8.

(iii) We have the prime factors of 10648 as:

\(\qquad \begin{array}{c|lcr} 2 & 10648\\ \hline 2 & 5324\\ \hline 2 & 2662\\ \hline 11 & 1331\\ \hline 11 & 121\\ \hline 11 & 11\\ \hline & 1\\ \end{array} \)

\(10648 =2 \times 2 \times 2 \times 11 \times 11 \times 11 = 2^3 \times 11^3 \times \)

∴\(\sqrt[3]{10648}=2 \times 11=22\)

Hence, the cube root of 10648 is 22.

(iv) We have the prime factors of 27000 as:

\(\qquad \begin{array}{c|lcr} 2 & 27000\\ \hline 2 & 13500\\ \hline 2 & 6750\\ \hline 3 & 3375\\ \hline 3 & 1125\\ \hline 3 & 375\\ \hline 5 & 125\\ \hline 5 & 25\\ \hline 5 & 5\\ \hline & 1\\ \end{array} \)

\(27000 =2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 = 2^3 \times 3^3 \times 5^3\)

∴\(\sqrt[3]{27000}=2 \times 3 \times 5=30\)

Hence, the cube root of 27000 is 30.

(v) We have the prime factors of 15625 as:

\(\qquad \begin{array}{c|lcr} 5 & 15625\\ \hline 5 & 3125\\ \hline 5 & 625\\ \hline 5 & 125\\ \hline 5 & 25\\ \hline 5 & 5\\ \hline & 1\\ \end{array} \)

\(15625 =5 \times 5 \times 5 \times 5 \times 5 \times 5= 5^3 \times 5^3\)

∴\(\sqrt[3]{15625}=5 \times 5=25\)

Hence, the cube root of 15625 is 25.

(vi) We have the prime factors of 13824 as:

\(\qquad \begin{array}{c|lcr} 2 & 13824\\ \hline 2 & 6912\\ \hline 2 & 3456\\ \hline 2 & 1728\\ \hline 2 & 864\\ \hline 2 & 432\\ \hline 2 & 216\\ \hline 2 & 108\\ \hline 2 & 54\\ \hline 3 & 27\\ \hline 3 & 9\\ \hline 3 &3 \\ \hline & 1\\ \end{array} \)

\(13824=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \)

\(\Rightarrow 2^3 \times 2^3 \times 2^3 \times 3^3\)

∴\(\sqrt[3]{13824}=2 \times 2 \times 2 \times 3=24\)

Hence, the cube root of 13824 is 24.

(vii) We have the prime factors of 110592 as:

\(\qquad \begin{array}{c|lcr} 2 & 110592\\ \hline 2 & 55296\\ \hline 2 & 27648\\ \hline 2 & 13824\\ \hline 2 & 6912\\ \hline 2 & 3456\\ \hline 2 & 1728\\ \hline 2 & 864\\ \hline 2 & 432\\ \hline 2 & 216\\ \hline 2 & 108\\ \hline 2 & 54\\ \hline 3 & 27\\ \hline 3 & 9\\ \hline 3 &3 \\ \hline & 1\\ \end{array} \)

\(110592=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \)

\(\Rightarrow 2^3 \times 2^3 \times 2^3 \times 2^3 \times 3^3\)

∴\(\sqrt[3]{110592}=2 \times 2 \times 2 \times 2 \times 3=48\)

Hence, the cube root of 110592 is 48.

(viii) We have the prime factors of 46656 as:

\(\qquad \begin{array}{c|lcr} 2 & 46656\\ \hline 2 & 23328\\ \hline 2 & 11664\\ \hline 2 & 5832\\ \hline 2 & 2916\\ \hline 2 & 1458\\ \hline 3 & 729\\ \hline 3 & 243\\ \hline 3 & 81\\ \hline 3 & 27\\ \hline 3 & 9\\ \hline 3 &3 \\ \hline & 1\\ \end{array} \)

\(46656=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)

\(\Rightarrow 2^3 \times 2^3 \times 3^3 \times 3^3\)

∴\(\sqrt[3]{46656}=2 \times 2 \times 3 \times 3=36\)

Hence, the cube root of 46656 is 36.

(ix) We have the prime factors of 175616 as:

\(\qquad \begin{array}{c|lcr} 2 & 175616\\ \hline 2 & 87808\\ \hline 2 & 43904\\ \hline 2 & 21952\\ \hline 2 & 10976\\ \hline 2 & 5488\\ \hline 2 & 2744\\ \hline 2 & 1372\\ \hline 2 & 686\\ \hline 7 & 343\\ \hline 7 & 49\\ \hline 7 &7 \\ \hline & 1\\ \end{array} \)

\(175616=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7 \)

\(\Rightarrow 2^3 \times 2^3 \times 2^3 \times 7^3\)

∴\(\sqrt[3]{175616}=2 \times 2 \times 2 \times 7 =56\)

Hence, the cube root of 175616 is 56.

(x) We have the prime factors of 91125 as:

\(\qquad \begin{array}{c|lcr} 3 & 91125\\ \hline 3 & 30375\\ \hline 3 & 10125\\ \hline 3 & 3375\\ \hline 3 & 1125\\ \hline 3 & 375\\ \hline 5 & 125\\ \hline 5 & 25\\ \hline 5 & 5\\ \hline & 1\\ \end{array} \)

\(91125 =3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 = 3^3 \times 3^3 \times 5^3\)

∴\(\sqrt[3]{91125}=3 \times 3 \times 5=45\)

Hence, the cube root of 91125 is 45.

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