(3x - 2y) : (3x + 3y) = 5 : 6, then one of the value of ${(\frac{\sqrt[3]{x} + \sqrt[3]{y}}{\sqrt[3]{x} - \sqrt[3]{y}})}^{2}$ ${\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2}$ is?

1 $\frac{1}{25}$ $\frac{1}{{25}}$
25
3 $\frac{1}{5}$ $\frac{1}{5}$
425

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{(3 x - 2 y)}{(3 x + 3 y)} = \frac{5}{6} \\ \Rightarrow 18 x - 12 y = 10 x + 15 y \\ \Rightarrow 8 x = 27 y \\ \Rightarrow \frac{x}{y} = \frac{27}{8} \\ ∴ {(\frac{\sqrt[3]{x} + \sqrt[3]{y}}{\sqrt[3]{x} - \sqrt[3]{y}})}^{2} \\ \Rightarrow {(\frac{\sqrt[3]{27} + \sqrt[3]{8}}{\sqrt[3]{27} - \sqrt[3]{8}})}^{2} \\ \Rightarrow {(\frac{3 + 2}{3 - 2})}^{2} \\ \Rightarrow {(5)}^{2} \\ \Rightarrow 25 \end{aligned}

$\eqalign{ & \frac{{\left( {3x - 2y} \right)}}{{\left( {3x + 3y} \right)}} = \frac{5}{6} \cr & \Rightarrow 18x - 12y = 10x + 15y \cr & \Rightarrow 8x = 27y \cr & \Rightarrow \frac{x}{y} = \frac{{27}}{8} \cr & \therefore {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr & \Rightarrow {\left( 5 \right)^2} \cr & \Rightarrow 25 \cr}$