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State and prove Euler's theorem for three variables and hence find the following

$x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z}$ where $u \;=\; \dfrac{x^3y^3z^3}{x^3+y^3+z^3}$

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Statement: If u=f(x, y, z)is a homogeneous function of degree n, then -

$\\ x \dfrac{\partial u}{\partial x} + y \dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z} \;=\; n \cdot u \\ \; \\ \; \\$

Let, u=f(x, y, z) is a homogeneous function of degree n.

$\therefore u \;=\; x^n f( \frac{y}{x}, \frac{z}{x} ) \; \; \ldots (i) \\ \; \\ \; \\$

Differentiate u partially w.r.t.x,we get-

$\\ \dfrac{\partial u}{\partial x} \;=\; nx^{n-1}f( \frac{y}{x}, \frac{z}{x} ) + x^n \dfrac{\partial}{\partial x}f( \frac{y}{x}, \frac{z}{x} ) \; \; \; \ldots (ii) \\$

Similarly,

$\\ \dfrac{\partial u}{\partial y} \;=\; x^n \dfrac{\partial}{\partial y}f( \frac{y}{x}, \frac{z}{x} ) \; \; \; \ldots (iii) \\ and \\ \dfrac{\partial u}{\partial z} \;=\; x^n \dfrac{\partial}{\partial z}f( \frac{y}{x}, \frac{z}{x} ) \; \; \; \ldots (iv) \\ \; \\ \; \\$

Now, let $\dfrac{y}{x}=v \; and \; \dfrac{z}{x}=w$

$\therefore \dfrac{\partial v}{\partial x} \;=\; \dfrac{-y}{x^2} \;,\; \\ \dfrac{\partial v}{\partial y} \;=\; \dfrac{1}{x} \\ \dfrac{\partial v}{\partial z} \;=\; 0 \\ \; \\ \; \\ \dfrac{\partial w}{\partial x} \;=\; \dfrac{-z}{x^2} \;,\; \\ \dfrac{\partial w}{\partial y} \;=\; 0 \\ \dfrac{\partial w}{\partial z} \;=\; \dfrac{1}{x} \\ \; \\$

Now, $\dfrac{\partial f}{\partial x} \;=\; \dfrac{\partial f}{\partial v} \cdot \dfrac{\partial v}{\partial x} + \dfrac{\partial f}{\partial w} \cdot \dfrac{\partial w}{\partial x} \\ \; \\ \; \\$

$\therefore \dfrac{\partial f}{\partial x} \;=\; \dfrac{\partial f}{\partial v} (\dfrac{-y}{x^2}) + \dfrac{\partial f}{\partial w} (\dfrac{-w}{x^2}) \; \; \ldots (v) \\ \; \\ \; \\$

Also, $\dfrac{\partial f}{\partial y} \;=\; \dfrac{\partial f}{\partial v} \cdot \dfrac{\partial v}{\partial y} + \dfrac{\partial f}{\partial w} \cdot \dfrac{\partial w}{\partial y} \\ \; \\ \; \\$

$\therefore \dfrac{\partial f}{\partial y} \;=\; \dfrac{\partial f}{\partial v} (\dfrac{1}{x}) + \dfrac{\partial f}{\partial w} (0) \;=\; \dfrac{1}{x} \dfrac{\partial f}{\partial v} \; \; \ldots (vi) \\ \; \\ \; \\$

And, $\dfrac{\partial f}{\partial z} \;=\; \dfrac{\partial f}{\partial v} \cdot \dfrac{\partial v}{\partial z} + \dfrac{\partial f}{\partial w} \cdot \dfrac{\partial w}{\partial z} \\ \; \\ \; \\$

$\therefore \dfrac{\partial f}{\partial z} \;=\; \dfrac{\partial f}{\partial v} (0) + \dfrac{\partial f}{\partial w} (\dfrac{1}{x}) \;=\; \dfrac{1}{x} \dfrac{\partial f}{\partial w} \; \; \ldots (vii) \\ \; \\ \; \\$

Substituting (v), (vi) and (vii) in (ii), (iii) and (iv) respectively,

$\\ \therefore \dfrac{\partial u}{\partial x} = nx^{n-1} ( \frac{y}{x}, \frac{z}{x} ) + x^n \Big[ (\dfrac{-y}{x^2})\dfrac{\partial f}{\partial v} + (\dfrac{-z}{x^2}) \dfrac{\partial f}{\partial w} \Big] \; \; \ldots (viii) \\ \; \\ \; \\ \; \\ \therefore \dfrac{\partial u}{\partial y} = nx^{n} \Big[ \dfrac{1}{x} \cdot \dfrac{\partial f}{\partial v} \Big] \; \; \ldots (ix) \\ \; \\ \; \\ \; \\ \therefore \dfrac{\partial u}{\partial z} = nx^{n} \Big[ \dfrac{1}{x} \cdot \dfrac{\partial f}{\partial w} \Big] \; \; \ldots (x) \\ \; \\ \; \\$

Now

$\\ x\dfrac{\partial u}{\partial x}+ y\dfrac{\partial u}{\partial y}+ z\dfrac{\partial u}{\partial z} \\ = x \bigg\{ nx^{n-1} ( \frac{y}{x}, \frac{z}{x} ) - x^n (\dfrac{y}{x^2})\dfrac{\partial f}{\partial v} - x^n (\dfrac{z}{x^2}) \dfrac{\partial f}{\partial w} \bigg\} \\ + y \bigg\{ nx^{n} \Big[ \dfrac{1}{x} \cdot \dfrac{\partial f}{\partial v} \Big] \bigg\} + z \bigg\{ nx^{n} \Big[ \dfrac{1}{x} \cdot \dfrac{\partial f}{\partial w} \Big] \bigg\} \\ \; \\ \; \\ \; \\ \; \\ \therefore x\dfrac{\partial u}{\partial x}+ y\dfrac{\partial u}{\partial y}+ z\dfrac{\partial u}{\partial z} \\ = nx^{n} ( \frac{y}{x}, \frac{z}{x} ) - x^n (\dfrac{y}{x})\dfrac{\partial f}{\partial v} - x^n (\dfrac{z}{x}) \dfrac{\partial f}{\partial w}+ \\ nx^{n} \dfrac{y}{x} \cdot \dfrac{\partial f}{\partial v}+ nx^{n} \dfrac{z}{x} \cdot \dfrac{\partial f}{\partial w} \\ \; \\ \; \\ \; \\$

$\therefore x\dfrac{\partial u}{\partial x}+ y\dfrac{\partial u}{\partial y}+ z\dfrac{\partial u}{\partial z} \;=\; nx^n( \frac{y}{x}, \frac{z}{x} ) \;=\; nu$ Hence Proved.

$\\ \; \\ \; \\ \; \\$

Now,$u \;=\; \dfrac{x^3y^3z^3}{x^3+y^3+z^3} \;=\; x^6 \dfrac{(\frac{y}{x})^3(\frac{z}{x})^3}{1+(\frac{z}{x})^3+(\frac{z}{x})^3} \;=\; x^6 f(\frac{y}{x},\frac{z}{x} ) \\ \; \\$

Hence, u is a homogeneous function of degree 6. Hence, by Euler’s theorem,

$x\dfrac{\partial u}{\partial x}+ y\dfrac{\partial u}{\partial y}+ z\dfrac{\partial u}{\partial z} \;=\; nu \;=\; 6u \;=\; \dfrac{6x^3y^3z^3}{x^3+y^3+z^3}$

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thanks for providing solution