(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.