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

$=5.646-5.540=0.106$

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

$=\frac{0.106}{10}\times 5=0.053$

$\therefore$ $\sqrt[3]{175}=5.540+0.053=5.593$

Now

$37800=6\times \sqrt[3]{175}=6\times 5.593=33.558$

Thus, the required cube root is 33.558.