Question

Making use of the cube root table, find the cube roots of
the following (correct to three decimal places).

$7342$

Answer 1

Step: Find the cube root of $7342$

$x$	$\sqrt[3]{3x}$	$\sqrt[3]{310x}$	$\sqrt[3]{3100x}$
$10$	$2.154$	$4.642$	$10$
$73$	$4.179$	$9.004$	$19.40$
$74$	$4.198$	$9.045$	$19.49$

$\because 7342=73.42\times 100$

$\therefore \sqrt[3]{37342}=\sqrt[3]{373.42\times 100}$

$=\sqrt[3]{3100}\times \sqrt[3]{373.42}$

$=4.642\times \sqrt[3]{373.42}$ $[\because \sqrt[3]{3100=4.642}]$

Now, $73.42$ lies between $73$ and $74$.
From the table,

$\sqrt[3]{373}=4.179\text{and}\sqrt[3]{374}=4.198$

$\therefore \text{Difference}=4.198–4.179=0.019$

For the difference of $(74-73)$ i.e., $1$, the difference in the values $=0.019$

$\therefore$ For the difference of $(73.42-73)$ i.e., 0.42, the difference in the value $=0.019\times 0.42=0.00798$

$\therefore \sqrt[3]{373.42}=4.179+0.00798$

$=4.18698$

$\therefore \sqrt[3]{37342}=4.642\times 4.18698=19.43596116$

$\approx 19.436$

Hence, the cube root of $7342$ is $19.436$.