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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Updated on: 10-Oct-2022

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