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Making use of the cube root table, find the cube root of the following (currect to three decimal places): 

(i) 7800

(ii) 1346

(iii) 250

(iv) 5112

1 Answer

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(i) 7800

7800 = 78×100

∴ \(\sqrt[3]{7800} = \sqrt[3]{78\times100}\) = \(\sqrt[3]{78}\times{\sqrt[3]{100}}\)

By using cube root table, 

We get,

\(\sqrt[3]{78} = 4.273 \) and \(\sqrt[3]{100} = 4.642\)

∴ \(\sqrt[3]{7800} = \sqrt[3]{78}\) = \(4.273\times4.642 = 19.835\)

(ii) 1346

By primefactorisation method, 

We get, 

1346 = 2×673

\(\sqrt[3]{1346} = \sqrt[3]{2}\times \)\(\sqrt[3]{673}\)

Also, 

670<673<680=>\(\sqrt[3]{670}<\sqrt[3]{673}<\sqrt[3]{680}\)

By using cube root table,

\(\sqrt[3]{670} = 8.750\) and \(\sqrt[3]{680} = 8.794\)

For the difference (680-670) which is 10. 

The difference in the values, 

= 8.794 - 8.750 = 0.044 

For the difference (673-670) which is 3. 

The difference in the values,

\(\frac{0.044}{10}\times3 = 0.0132 = 0.013\)

\(\sqrt[3]{673} = 8.750 + 0.013 = 8.763\)

so,

\(\sqrt[3]{1346} = \sqrt[3]{2}\times\sqrt[3]{673} = 1.260\times8.763 = 11.041\)

(iii) 250

250 = 25×100 By using cube root table 250 would be in column \(\sqrt[3]{10x}\) against 25.

So we get,

\(\sqrt[3]{250} = 6.3\)

(iv) 5112

\(\sqrt[3]{5112}\) = \(\sqrt[3]{2\times2\times2\times3\times3\times71}\) = \(\sqrt[3]{2^3\times3^2\times71}\) = \(2\times\sqrt[3]{9}\times\sqrt[3]{71}\)

From cube root table we get,

\(\sqrt[3]{9} = 2.080\) and \(\sqrt[3]{71} = 4.141\)

Hence,

\(\sqrt[3]{5112} = 2\times2.080\times4.141 = 17.227\)

Thus, 

the required cube root is = 17.227.

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