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If $( \frac{( 3 x-4)^{3}-( x+1)^{3}}{( 3 x-4)^{3}+( x+1)^{3}}=\frac{61}{189})$, find the value of $x$.
Given: $\frac{( 3x-4)^{3}-( x+1)^{3}}{( 3x-4)^{3}+( x+1)^{3}}=\frac{61}{189}$.
To do: To find the value of $x$.
Solution:
$\frac{( 3x-4)^{3}-( x+1)^{3}}{( 3x-4)^{3}+( x+1)^{3}}=\frac{61}{189}$
$\Rightarrow 189( 3x-4)^{3} -189( x+1)^{3} = 61( 3x-4)^{3} + 61( x+1)^{3}$
$\Rightarrow ( 189 - 61)( 3x-4)^{3} = ( 61 + 189)( x+1)^{3}$
$\Rightarrow 128( 3x-4)^{3} = 250( x+1)^{3}$
$\Rightarrow 64( 3x-4)^{3} = 125( x+1)^{3}$
$\Rightarrow 4^{3}( 3x-4)^{3} = 5^{3}( x+1)^{3}$ [Taking cube root of both sides]
$\Rightarrow 4( 3x - 4) = 5( x + 1)$
$\Rightarrow 12x - 16 = 5x + 5$
$\Rightarrow 7x = 21$
$\Rightarrow x = 3$
Thus, $x=3$
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