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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 $\frac{29}{5}$.
\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}