Question

Making use of the cube root table, find the cube root
37800 .

Answer 1

We have: \[37800 = 2^3 \times 3^3 \times 175 \Rightarrow \sqrt[3]{37800} = \sqrt[3]{2^3 \times 3^3 \times 175} = 6 \times \sqrt[3]{175}\]

Also

\[170 < 175 < 180 \Rightarrow \sqrt[3]{170} < \sqrt[3]{175} < \sqrt[3]{180}\]

From cube root table, we have: \[\sqrt[3]{170} = 5 . 540 \text{ and }  \sqrt[3]{180} = 5 . 646\]

For the difference (180 - 170), i.e., 10, the difference in values

\[= 5 . 646 - 5 . 540 = 0 . 106\]

∴  For the difference of (175 - 170), i.e., 5, the difference in values

\[= \frac{0 . 106}{10} \times 5 = 0 . 053\]

∴ \[\sqrt[3]{175} = 5 . 540 + 0 . 053 = 5 . 593\]

Now 

\[37800 = 6 \times \sqrt[3]{175} = 6 \times 5 . 593 = 33 . 558\]

Thus, the required cube root is 33.558.