If \(\sqrt{x} \sqrt[3]{y} = (x + y)^n\) and \(x\dfrac{dy}{dx} - y = 0\), then \(n = \)

Given equation: \(\sqrt{x} \sqrt[3]{y} = (x + y)^n\)

Taking natural logarithm on both sides:

\(\dfrac{1}{2} \ln x + \dfrac{1}{3} \ln y = n \ln(x + y)\)

Differentiating with respect to \(x\):

\(\dfrac{1}{2x} + \dfrac{1}{3y} \dfrac{dy}{dx} = \dfrac{n}{x + y} \left(1 + \dfrac{dy}{dx}\right)\)

It is given that \(x \dfrac{dy}{dx} - y = 0 \Rightarrow \dfrac{dy}{dx} = \dfrac{y}{x}\)

Substituting \(\dfrac{dy}{dx} = \dfrac{y}{x}\) in the differentiated equation:

\(\dfrac{1}{2x} + \dfrac{1}{3y} \left(\dfrac{y}{x}\right) = \dfrac{n}{x + y} \left(1 + \dfrac{y}{x}\right)\)

\(\dfrac{1}{2x} + \dfrac{1}{3x} = \dfrac{n}{x + y} \left(\dfrac{x + y}{x}\right)\)

\(\dfrac{5}{6x} = \dfrac{n}{x}\)

\(n = \dfrac{5}{6}\)

Answer: \(\dfrac{5}{6}\)