Theory:

1. Find the cube of $$21$$.

Solution:

If you multiply a number by itself and then by itself again (thrice), the result is a cube number.

$$21 = 21 \times 21 \times 21 = 9261$$

Therefore, the cube of $$21$$ is $$9261$$.

2. Is $$625$$ is a perfect cube? If not, make it as a perfect cube by the multiplication of smallest number.

Solution:

$\begin{array}{l}\underset{¯}{5|625\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}}\\ \underset{¯}{5|125\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}}\\ \underset{¯}{5|25\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}}\\ \underset{¯}{5|5\phantom{\rule{0.294em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.294em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}}\\ \phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}|\phantom{\rule{0.147em}{0ex}}1\end{array}$

Group the factors in pair of three numbers.

$$125 = (5 \times 5 \times 5) \times 5$$

Here, factor $$5$$ is leftover.

So, $$625$$ is not a perfect cube.

To make it as a perfect cube, we need two more $$5$$'s.

So, multiply the given number by two $$5$$'s.

$$625 \times 5 \times 5 = 15625$$

Therefore, $$15625$$ is a perfect cube.

3. Is $$243$$ is a perfect cube? If not, find the smallest number to divide $$243$$ to make it as a perfect cube.

Solution:

$$243 = (3 \times 3 \times 3) \times 3 \times 3$$

Here, factor $$3 \times 3 = 9$$ is leftover while grouping.

So, $$243$$ is not a perfect cube.

To make it as a perfect cube, divide the given number by the leftover factor $$9$$.

$\frac{243}{9}=27$

Therefore, $$9$$ is the smallest number to divide $$243$$, and the obtained perfect square is $$27$$.