If (3x - 2y) : (2x + 3y) = 5 : 6,then one of | vedclass

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(B) Given the ratio: $\frac{3x - 2y}{2x + 3y} = \frac{5}{6}$Cross-multiplying gives: $6(3x - 2y) = 5(2x + 3y)$$18x - 12y = 10x + 15y$$18x - 10x = 15y

Vedclass · Accepted Answer

$25$

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(B) Given the ratio: $\frac{3x - 2y}{2x + 3y} = \frac{5}{6}$Cross-multiplying gives: $6(3x - 2y) = 5(2x + 3y)$$18x - 12y = 10x + 15y$$18x - 10x = 15y + 12y$$8x = 27y$$\frac{x}{y} = \frac{27}{8}$Taking the cube root on both sides: $\frac{\sqrt[3]{x}}{\sqrt[3]{y}} = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}$Applying the componendo and dividendo rule: $\frac{\sqrt[3]{x} + \sqrt[3]{y}}{\sqrt[3]{x} - \sqrt[3]{y}} = \frac{3 + 2}{3 - 2} = \frac{5}{1} = 5$Squaring both sides: $\left(\frac{\sqrt[3]{x} + \sqrt[3]{y}}{\sqrt[3]{x} - \sqrt[3]{y}}\right)^{2} = 5^{2} = 25$

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

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