Question:
Making use of the cube root table, find the cube root
37800
Solution:
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:
$\sqrt[3]{170}=5.540$ 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.