Ncert solutions

Grade 8

# Question 1

Q1) Find the cube root of each of the following numbers by prime factorization method:

(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

Solution:

(i) 64

\sqrt[3]{64}=\sqrt[3]{2\times2\times2\times2\times2\times2}

\sqrt[3]{64}=\ 2\times2

= 4

(ii) 512

\sqrt[3]{512}=\sqrt[3]{2\times2\times2\times2\times2\times2\times2\times2\times2}

= 2 x 2 x 2 = 8

(iii) 10648

\sqrt[3]{10648}=\sqrt[3]{2\times2\times2\times11\times11\times11}

= 2 x 11

= 22

(iv) 27000

\sqrt[3]{27000}=\sqrt[3]{2\times2\times2\times3\times3\times3\times5\times5\times5}

= 2 x 3 x 5

= 30

(v) 15625

\sqrt[3]{15625}=\sqrt[3]{5\times5\times5\times5\times5\times5}

= 5 x 5

= 25

(vi) 13824

\sqrt[3]{13824}=\sqrt[3]{2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3}

= 2 x 2 x 2 x 3

= 24

(vii) 110592

\sqrt[3]{110592}=\sqrt[3]{2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3}

= 2 x 2 x 2 x 2 x 3

= 48

(viii) 46656

\sqrt[3]{46645}=\sqrt[3]{2\times2\times2\times2\times2\times2\times3\times3\times3\times3\times3\times3}

= 2 x 2 x 3 x 3

= 36

(ix) 175616

\sqrt[3]{175616}=\sqrt[3]{2\times2\times2\times2\times2\times2\times2\times2\times2\times7\times7\times7}

= 2 x 2 x 2 x 7

= 56

(x) 91125

\sqrt[3]{91125}=\sqrt[3]{3\times3\times3\times3\times3\times3\times5\times5\times5}

= 3 x 3 x 5 = 45

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