Question

Find the cube root of each of the following numbers by prime Factorization method:

(i) 729 (ii) 343 (iii) 512 (iv) 0.064
(v) 0.216 (vi) 5 23/64 (vii) – 1.331 (viii) – 27000

Answer 1

i) 9
ii) 7
iii) 8
iv) 0.4
v) 0.6
vi)
vii) -11
viii) -30

Answer 2

Answer:

We have to find cube root of the given numbers

First we have to factorize the given numbers:

$(i)\sqrt[3]{729} =\sqrt[3]{9\times 9 \times 9} = \sqrt[3]{9^3} = 9\\(ii) \sqrt[3]{343}=\sqrt[3]{7\times 7 \times 7} = \sqrt[3]{7^3} = 7\\(iii) \sqrt[3]{512}= \sqrt[3]{8\times8 \times 8} = \sqrt[3]{8^3} = 8\\(iv) \sqrt[3]{0.064} = \sqrt[3]{\dfrac {64}{1000}} = { \sqrt[3]{\dfrac {4\times 4 \times 4}{10\times 10 \times 10}}}\\\sqrt[3]{0.064}= \dfrac 4{10}= 0.4$

$(v) \sqrt[3]{0.216} = \sqrt[3]{\dfrac {216}{1000}} = { \sqrt[3]{\dfrac {6\times 6 \times 6}{10\times 10 \times 10}}}\\ \sqrt[3]{0.216}= \dfrac 6{10}} = 0.6\\\\(vi) \sqrt[3]{5\dfrac {23}{64}}= \sqrt[3]{\dfrac{343}{64} } = \sqrt[3]{\dfrac{7\times 7 \times 7}{4\times 4 \times 4}} = \dfrac{7}{4} =1\dfrac{3}{4}\\(vii) -\sqrt[3]{1.331} = -\sqrt[3]{\dfrac {1331}{1000}} =-{ \sqrt[3]{\dfrac {11\times 11 \times 11}{10\times 10 \times 10}}}\\ -\sqrt[3]{1.331}=-\dfrac {11}{10}} = -1.1$

$(viii) \sqrt[3]{-27000}= -\sqrt[3]{30\times 30 \times 30} = -\sqrt[3]{{30}^3} = -30$