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$$ u=x^{3}+3 x^{2} y+3 x y^{2}+y^{3} \text {, prove that } x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 u \text {. } $$
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Solution:

$$ \begin{aligned} u &=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}.....(A) \\\\ \frac{\partial u}{\partial x} &=3 x^{2}+6 x y+3 y^{2}+0 \\\\ x \frac{\partial u}{\partial x} &=x\left(3 x^{2}+6 x y+3 y^{2}\right) \\\\ x \frac{\partial u}{\partial x} &=3 x^{3}+6 x^{2} y+3 x y^{2}........(1) \\\\ u &=x^{3}+3 x^{2} y+3 x y^{2}+y^{3} \\\\ \frac{\partial u}{\partial y} &=0+3 x^{2}+3 x(2 y)+3 y^{2} \\\\ &=3 x^{2}+6 x y+3 y^{2} \\\\ y \frac{\partial u}{\partial y} &=y\left(3 x^{2}+6 x y+3 y^{2}\right) \\\\ &=3 x^{2} y+6 x y^{2}+3 y^{3}.....(2) \\\\ (1)+(2) & \Rightarrow \\\\ \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y} &=3 x^{3}+6 x^{2} y+3 x y^{2}+3 x^{2} y+6 x y^{2}+3 y^{3} \\\\ &=3 x^{3}+9 x^{2} y+9 x y^{2}+3 y^{3} \\\\ &=3\left(x^{3}+3 x^{2} y+3 x y^{2}+y^{3}\right) \\\\ &=3u\ using(A) \end{aligned} $$

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