Which is the smallest number by which 725 must be divided to make it a perfect cube?

Given :

The given number is 725.

To do :

We have to find the smallest number by which 725 must be divided to make it a perfect cube.

Solution :

To find the smallest number by which 725 must be divided to make it a perfect cube, we have to find the prime factors of it.

Prime factorization of 725 is,

$725=5\times 5\times 29$

If we divide the given number by 29 it will be a perfect square.

Therefore, for 725 to be a perfect cube we have to divide 725 by $\frac{29}{5}$.

$\frac{725}{\frac{29}{5}} = \frac{5\times 5\times 29}{\frac{29}{5}}$

$\frac{725 \times 5}{29} = \frac{5\times 5\times 29 \times 5}{29}$

$125 = 5\times 5\times 5 = 5^3$

Therefore, the smallest number by which 725 must be divided to make it a perfect cube is $\frac{29}{5}$.

$\begin{array}{l}$
\frac{725}{\frac{29}{5}} =\frac{5\times 5\times 29}{\frac{29}{5}}\
\
\frac{725\times 5}{29} =\frac{5\times 5\times 29\times 5}{29}\
\
125=5\times 5\times 5=5^{3}
\end{array}