If the cube of $1+\frac{3}{5}$ is $4+\frac{x}{125}$, find $x$.
Given: The cube of $( 1+\frac{3}{5})$ is $( 4+\frac{x}{125})$.
To do: To find the $x$.
Solution:
As given, The cube of $( 1+\frac{3}{5})$ is$( 4+\frac{x}{125})$
$\Rightarrow ( 1+\frac{3}{5})^3=( 4+\frac{x}{125})$
$\Rightarrow ( \frac{5+3}{5})^3=( 4+\frac{x}{125})$
$\Rightarrow \frac{8}{5}\times\frac{8}{5}\times\frac{8}{5}=4+\frac{x}{125}$
$\Rightarrow \frac{8\times8\times8}{5\times5\times5}=4+\frac{x}{125}$
$\Rightarrow \frac{512}{125}=4+\frac{x}{125}$
$\Rightarrow \frac{512}{125}-4=\frac{x}{125}$
$\Rightarrow \frac{512-500}{125}=\frac{x}{125}$
$\Rightarrow \frac{12}{125}=\frac{x}{125}$
$\Rightarrow x=12$
Thus, $x=12$.
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