Jacobian
If \(u\) and \(v\) are the functions of two independent variables \(x\) and \(y\) then Jacobian of \(u\) and \(v\) w.r.t. \(x, y\) is: \[ J = J\left(\frac{u,v}{x,y} \right) = \frac{\partial (u,v)}{\partial (x,y)} \] \[ = \Large \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} \] Similarly, for three variables, \[ J = J\left(\frac{u,v,w}{x,y,z} \right) = \frac{\partial (u,v,w)}{\partial (x,y,z)} \] \[ = \Large \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{vmatrix} \] Applications : (1) In change of variables in Multiple Intgration. Example: changing Cartesian coordinates \(x,y\) to Polar coordinates \(r, \theta \) in Double Integral \[ x = r\cos{\theta}, y = r\sin{\theta} \] \[ \text{In } \int xydxdy \longrightarrow dxdy = Jdrd\theta \] (2) To find Functional Relationship between two variables
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Properties :
(1) If \(u\) and \(v\) are functions of \(x\) and \(y\) then
\[ J = J\left(\frac{u.v}{x,y} \right) = \frac{\partial (u,v)}{\partial (x,y)} \]
\[ J' = J\left(\frac{x.y}{u,v} \right) = \frac{\partial (x,y)}{\partial (u,v)} \]
\[ J \times J' = 1 \]
(2) Chain Rule : If If \(u\) and \(v\) are functions of \(x\) and \(y\) and \(x\) and \(y\) are functions of \(r\) and \(s\) then
\[ J\left(\frac{u.v}{r,s} \right) = J\left(\frac{u.v}{x,y} \right) \times J\left(\frac{x,y}{r,s} \right) \]
\[ \frac{\partial (u,v)}{\partial (r,s)} = \frac{\partial (u,v)}{\partial (x,y)}\times \frac{\partial (x,y)}{\partial (r,s)} \]
Sample Problem 1 :
If \(x= r\cos{\theta}, y = r\sin{\theta} \), then show that \( J\Large \left(\Large \frac{x,y}{r,\theta} \right) = r \)
Solution :
\[ J\left(\frac{x,y}{r,\theta} \right) = r \] \[ = \Large \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} \] Partially differentiating \(x,y\) w.r.t. \(r,\theta \) \[ \frac{\partial x}{\partial r} = \cos{\theta}, \hspace{30pt} \frac{\partial y}{\partial r} = \sin{\theta} \] \[ \frac{\partial x}{\partial \theta} = -r\sin{\theta}, \hspace{30pt} \frac{\partial y}{\partial \theta} = r\cos{\theta} \] Using Jacobian formula, \[ \begin{vmatrix} \cos{\theta} & -r\sin{\theta} \\ \sin{\theta} & r\cos{\theta} \end{vmatrix} \] \[ = r\cos^2{\theta} + r\sin^2{\theta} \] \[ = r \] Hence Proved
Sample Problem 2 :
If \( \large u = \Large \frac{yz}{x}, v = \Large \frac{zx}{y}, w = \Large \frac{xy}{z} \), then find \( \Large \frac{\partial (u,v,w)}{\partial (x,y,z)} \)
Solution :
Partially differentiating equations w.r.t. \(x,y,z\) \[ \frac{\partial u}{\partial x} = \frac{-yz}{x^2}, \hspace{30pt} \frac{\partial u}{\partial y} = \frac{z}{x} , \hspace{30pt} \frac{\partial u}{\partial z} = \frac{y}{x} \] \[ \frac{\partial v}{\partial x} = \frac{z}{y}, \hspace{30pt} \frac{\partial v}{\partial y} = \frac{-zx}{y^2}, \hspace{30pt} \frac{\partial v}{\partial z} = \frac{x}{y} \] \[ \frac{\partial w}{\partial x} = \frac{y}{z}, \hspace{30pt} \frac{\partial w}{\partial y} = \frac{x}{z}, \hspace{30pt} \frac{\partial w}{\partial z} = \frac{-xy}{z^2} \] \[ J = \frac{\partial (u,v,w)}{\partial (x,y,z)} \] \[ = \Large \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{vmatrix} \] \[ = \Large \begin{vmatrix} \frac{-yz}{x^2} & \frac{z}{x} & \frac{y}{x} \\ \frac{z}{y} & \frac{-zx}{y^2} & \frac{x}{y} \\ \frac{y}{z} & \frac{x}{z} & \frac{-xy}{z^2} \end{vmatrix} \] \[ = \frac{-yz}{x^2}\left[\left(\frac{-zx}{y^2} \right)\left(\frac{-xy}{z^2} \right) - \left( \frac{x}{y} \right)\left(\frac{x}{z} \right) \right] \] \[ - \frac{z}{x}\left[\left(\frac{z}{y} \right)\left(\frac{-xy}{z^2} \right) - \left(\frac{y}{z}\right)\left(\frac{x}{y} \right) \right] \] \[ + \frac{y}{x}\left[\left(\frac{z}{y} \right)\left(\frac{x}{z} \right) + \left(\frac{zx}{y^2} \right)\left(\frac{y}{z} \right) \right] \] \[ = \frac{-yz}{x^2}\left[\frac{x^2}{yz} - \frac{x^2}{yz} \right] - \frac{z}{x}\left[\frac{-zx}{z} \right] + \frac{y}{x}\left[\frac{x}{y} + \frac{x}{y} \right] \] \[ = 4 \]
Sample Problem 3 :
If \( x = r\cos{\theta}, y = r\sin{\theta} \) then prove that \( \Large \frac{\partial (x,y)}{\partial (r,\theta)} \times \frac{\partial (r,\theta)}{\partial (x,y)} = 1 \)
Solution :
\[ \frac{\partial (x,y)}{\partial (r,\theta)} = \Large \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} \] \[ = \begin{vmatrix} \cos{\theta} & -r\sin{\theta} \\ \sin{\theta} & r\cos{\theta} \end{vmatrix} \] \[ = r\cos^2{\theta} + r\sin^2{\theta} \] \[ = r \] \[ \frac{\partial (r,\theta)}{\partial (x,y)} = \Large \begin{vmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y} \end{vmatrix} \] \[ x = r\cos{\theta} \Rightarrow r = x\sec{\theta} \] \[ y = r\sin{\theta} \Rightarrow r = y\hspace{3pt}cosec\hspace{3pt} \theta \] \[ \frac{\partial r}{\partial x} = \sec{\theta} , \hspace{30pt} \frac{\partial r}{\partial y} = cosec\hspace{3pt} \theta \] If you are thinking like that, then you are wrong, because \(r\) and \(\theta\) must be functions of same group of variables i.e. \(x\) and \(y\) just like \(x\) and \(y\) are functions of \(r\) and \(\theta\). That means we have to remove \(\theta\) from \(r\) and reduce \(r\) into function of \(x\) and \(y\). Similarly we will remove \(r\) from \(\theta\) and reduce \(\theta\) into function of \(x\) and \(y\). \[ x = r\cos{\theta} \Rightarrow \frac{x}{r} = \cos{\theta} \] \[ y = r\sin{\theta} \Rightarrow \frac{y}{r} = \sin{\theta} \] \[ \text{As we know, } \cos^2{\theta} + \sin^2{\theta} = 1 \] \[ \frac{x^2}{r^2} + \frac{y^2}{r^2} = 1 \] \[ x^2 + y^2 = r^2 \] \[ r = \sqrt{x^2 + y^2} \] \[ \text{Now, } \frac{\partial r}{\partial x} = \frac{1}{2}(x^2 + y^2)^\frac{1}{2}\times 2x \] \[ \Rightarrow \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \] \[ \text{Similarly, } \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \] \[ \text{Now for } \frac{\partial \theta}{\partial x} \text{ and } \frac{\partial \theta}{\partial y} \] \[ \frac{y}{x} = \frac{r\sin{\theta}}{r\cos{\theta}} \] \[ \Rightarrow \frac{y}{x} = \tan{\theta} \] \[ \Rightarrow \theta = \tan^{-1}{\frac{y}{x}} \] \[ \frac{\partial \theta}{\partial x} = \frac{1}{1 + \large \left(\frac{y}{x} \right)^2 }\left(\frac{-y}{x^2} \right) \] \[ \frac{\partial \theta}{\partial x} = \frac{-y}{x^2 + y^2} \] \[ \frac{\partial \theta}{\partial y} = \frac{1}{1 + \large \left(\frac{y}{x} \right)^2 }\left(\frac{1}{x} \right) \] \[ \frac{\partial \theta}{\partial y} = \frac{x}{x^2 + y^2} \] Now putting derivatives, \[ J = \Large \begin{vmatrix} \frac{x}{\sqrt{x^2 + y^2}} & \frac{y}{\sqrt{x^2 + y^2}} \\ \frac{-y}{x^2 + y^2} & \frac{x}{x^2 + y^2} \end{vmatrix} \] \[ = \left(\frac{x}{\sqrt{x^2 + y^2}} \right)\left(\frac{x}{x^2 + y^2} \right) - \left(\frac{-y}{x^2 + y^2} \right)\left(\frac{y}{\sqrt{x^2 + y^2}} \right) \] \[ = \frac{1}{\sqrt{x^2 + y^2}} \] \[ = \frac{1}{r} \] \[ \text{Now, } \frac{\partial (x,y)}{\partial (r,\theta)}\times \frac{\partial (r,\theta)}{\partial (x,y)} \] \[ = r \times \frac{1}{r} \] \[ = 1 \] = R.H.S.
Sample Problem 4 - Composite function :
If \(u = x^2 - y^2, v = 2xy \text{ and } x = x\cos{\theta}, y = r\sin{\theta} \), find \( J\Large \left(\frac{u,v}{r,\theta} \right) \)
Solution :
Here, as you can see, \(u\) and \(v\) are functions of \(x\) and \(y\), and \(x\) and \(y\) are functions of \(r\) and \(\theta\). So there are two methods to solve this problem:
Method 1 : Substitution
Method 2 : Use Chain Rule
\[ J\left(\frac{u,v}{r,\theta} \right) = J\left(\frac{u,v}{x,y} \right) \times J\left(\frac{x,y}{r,\theta} \right) \]
\[ \text{Now, } J\left(\frac{u,v}{x,y} \right) = \frac{\partial (u,v)}{\partial (x,y)} \]
\[ = \Large \begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial u}{\partial y}
\end{vmatrix}
\]
\[ = \begin{vmatrix}
2x & -2y \\
2y & 2x
\end{vmatrix}
\]
\[ = 4x^2 + 4y^2 \]
\[ J\left(\frac{x,y}{r,\theta} \right) = \Large \begin{vmatrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}
\end{vmatrix}
\]
\[
= \begin{vmatrix}
\cos{\theta} & -r\sin{\theta} \\
\sin{\theta} & r\cos{\theta}
\end{vmatrix}
\]
\[ = r(\cos^2{\theta} + \sin^2{\theta}) \]
\[ = r \]
Using Chain Rule
\[ J\left(\frac{u,v}{r,\theta} \right) = 4(x^2 + y^2)\times r \]
\[ = 4r(x^2 + y^2) \]
\[ = 4r(r^2\cos^2{\theta} + r^2\cos^2{\theta}) \]
\[ = 4r^3 \]
Sample Problem 5
If \(x = r\sin{\theta}\cos{\phi}, y = r\sin{\theta}\sin{\phi}\) and \( z = r\cos{\theta} \), then show that \( J = r^2\sin{\theta} \)
Solution :
\[ x = r\sin{\theta}\cos{\phi} \] \[ y = r\sin{\theta}\sin{\phi} \] \[ z = r\cos{\theta} \] Now, \[ J = J\left(\frac{x,y,z}{r,\theta,\phi} \right) \] \[ = \Large \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} \end{vmatrix} \] \[ = \begin{vmatrix} \sin{\theta}\cos{\phi} & r\cos{\theta}\cos{\phi} & -r\sin{\theta}\sin{\phi} \\ \sin{\theta}\sin{\phi} & r\cos{\theta}\sin{\phi} & r\sin{\theta}\cos{\phi} \\ \cos{\theta} & -r\sin{\theta} & 0 \end{vmatrix} \] Expanding along \(R_3\), \[ = \cos{\theta}[(r\sin{\theta}\cos{\phi})(r\cos{\theta}\cos{\phi}) - (r\cos{\theta}\sin{\phi})(-r\sin{\theta}\sin{\phi})] \] \[ - (-r\sin{\theta})[(\sin{\theta}\cos{\phi})(r\sin{\theta}\cos{\phi}) - (-r\sin{\theta}\sin{\phi})(\sin{\theta}\sin{\phi})] \] \[ + 0 \] \[ = \cos{\theta}(r^2\sin{\theta}\cos{\theta}\cos^2{\phi} + r^2\cos{\theta}\sin{\theta}\sin^2{\phi} ) \] \[ + r\sin{\theta}(r\sin^2{\theta}\cos^2{\phi} + r\sin^2{\theta}\sin^2{\phi}) \] \[ = \cos{\theta}[r^2\cos{\theta}\sin{\theta}(\cos^2{\phi} + \sin^2{\phi})] \] \[ + r\sin{\theta}[r\sin^2{\theta}(\cos^2{\phi} + \sin^2{\phi})] \] \[ = r^2\cos^2{\theta}\sin{\theta} + r^2\sin^3{\theta} \] \[ = r^2\sin{\theta}(\cos^2{\theta} + \sin^2{\theta}) \] \[ = r^2\sin{\theta} \] Hence Proved
Sample Problem 6 - Implicit Function (reducable to explicit)
If \(x + y + z = u; y + z = uv; z = uvw \), find \(J \Large \left(\frac{x,y,z}{u,v,w} \right) \)
Solution :
\[ x + y + z = u \hspace{25pt} \longrightarrow (1) \] \[ y + z = uv \hspace{25pt} \longrightarrow (2) \] \[ z = uvw \hspace{25pt} \longrightarrow (3) \] Using equation (3) in (2) \[ y + uvw = uv \] \[ \boxed{y = uv - uvw} \] Putting value of \(y + z\) in equation (1) \[ \boxed{x = u - uv} \] \[ \text{Now, } J\left(\frac{x,y,z}{u.v.w} \right) = \frac{\partial (x,y,z)}{\partial (u,v,w)} \] \[ = \Large \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{vmatrix} \] \[ = \begin{vmatrix} 1 - v & -u & 0 \\ v - vw & u - uw & -uv \\ vw & uw & uv \end{vmatrix} \] \[ = (1 - v)[(u - uw)uv + u^2vw] \] \[ + u[(u - vw)uv + uv^2w] \] \[ = (1 - v)(u^2v - \cancel{u^2vw} + \cancel{u^2vw}) + u(uv^2 - \cancel{uv^2w} + \cancel{uv^2w}) \] \[ = u^2v - \cancel{u^2v^2} + \cancel{u^2v^2} \] \[ = u^2v \]
Sample Problem 7 - Implicit function (not reducable to explicit) :
If \[ \begin{align*} u &= xyz \\ v &= x^2 + y^2 + z^2 \\ w &= x + y + z \end{align*} \] find \(J\Large \left(\frac{x,y,z}{u,v,w} \right) \)
Solution :
There are two ways to solve this:
Method 1 : Using formula \(J \times J' = 1 \Rightarrow \boxed{J' = \frac{1}{J}} \)
Method 2 : Using another formula, used especially for such problems
Method 1 is easy, so here we are showing Method 2:
\[ \begin{align*}
u = xyz &\Rightarrow xyz - u = 0 \\
v = x^2 + y^2 + z^2 &\Rightarrow x^2 + y^2 + z^2 - v = 0 \\
w = x + y + z &\Rightarrow x + y + z - w = 0
\end{align*}
\]
Let
\[ f_1 = xyz - u \]
\[ f_2 = x^2 + y^2 + z^2 - w \]
\[ f_3 = x + y + z - w \]
Using formula:
\[ J\left(\frac{x,y,z}{u,v,w} \right) = (-1)^n \Large \frac{J\left(\frac{f_1,f_2,f_3}{u,v,w} \right)}{J\left(\frac{f_1,f_2,f_3}{x,y,z} \right)} \]
\[\text{where } n = \text{ number of variables in one set, here } n = 3 \]
\[ \text{Now, } J\left(\frac{f_1,f_2,f_3}{x,y,z} \right) = \frac{\partial (f_1,f_2,f_3)}{\partial (x,y,z)} \]
\[ =
\Large \begin{vmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_2}{\partial y} & \frac{\partial f_1}{\partial z} \\
\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} & \frac{\partial f_2}{\partial z} \\
\frac{\partial f_3}{\partial x} & \frac{\partial f_3}{\partial y} & \frac{\partial f_3}{\partial z}
\end{vmatrix}
\]
\[
= \begin{vmatrix}
yz & xz & xy \\
2x & 2y & 2z \\
1 & 1 & 1
\end{vmatrix}
\]
\[ = yz(2y - 2z) - xz(2x - 2z) + xy(2x - 2y) \]
\[ = 2y^2z - 2yz^2 - 2x^2z + 2xz^2 + 2x^2y - 2xy^2 \]
\[ \text{Now, } J\left(\frac{f_1,f_2,f_3}{u,v,w} \right) \]
\[
=
\Large \begin{vmatrix}
\frac{\partial f_1}{\partial u} & \frac{\partial f_2}{\partial v} & \frac{\partial f_1}{\partial w} \\
\frac{\partial f_2}{\partial u} & \frac{\partial f_2}{\partial v} & \frac{\partial f_2}{\partial w} \\
\frac{\partial f_3}{\partial u} & \frac{\partial f_3}{\partial v} & \frac{\partial f_3}{\partial w}
\end{vmatrix}
\]
\[ =
\begin{vmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{vmatrix}
\]
\[ = -1 \]
\[ \text{Now, } J\left(\frac{x,y,z}{u,v,w} \right) = (-1)^3 \Large \frac{J\left(\frac{f_1,f_2,f_3}{u,v,w}
\right)}{J\left(\frac{f_1,f_2,f_3}{x,y,z} \right)} \]
\[ = \frac{(-1)(-1)}{2(x^2y - xy^2 + y^2z - yz^2 + xz^2 - x^2z)} \]
\[ = \frac{1}{2(x^2y - xy^2 + y^2z - yz^2 + xz^2 - x^2z)} \]
Functional Dependancy :
If \(u, v\) are functons of \(x, y\), then the necessary condition for the existance of functional dependancy is \[ J\left(\frac{u,v}{x,y} \right) = 0 \]
Sample Problem 8 :
Show that function \( u = \Large \frac{x + y}{1 - xy}\), \( v = \tan^{-1}{x} + \tan^{-1}{y} \) are dependant and find relationship between them.
Solution :
\[ u = \frac{x + y}{1 - xy} \] \[ \frac{\partial u}{\partial x} = \frac{(1 - xy)(1) - (x + y)(-y)}{(1 - xy)^2} \] \[ = \frac{1 - \cancel{xy} + \cancel{xy} - y^2}{(1 - xy)^2} \] \[ \frac{\partial u}{\partial x} = \frac{1 - y^2}{(1 - xy)^2} \] \[ \frac{\partial u}{\partial y} = \frac{(1 -xy)(1) - (x + y)(-x)}{(1 - xy)^2} \] \[ \frac{\partial u}{\partial y} = \frac{1 + x^2}{(1 - xy)^2} \] \[ \frac{\partial v}{\partial x} = \frac{1}{1 + x^2} \] \[ \frac{\partial v}{\partial y} = \frac{1}{1 + y^2} \] \[ J\left(\frac{u,v}{x,y} \right) = \Large \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} \] \[ = \Large \begin{vmatrix} \frac{1 + y^2}{(1 - xy)^2} & \frac{1 + x^2}{(1 - xy)^2} \\ \frac{1}{1 + x^2} & \frac{1}{1 + y^2} \end{vmatrix} \] \[ = \left[\frac{1 + y^2}{(1 - xy)^2} \right]\left[\frac{1}{1 + y^2} \right] - \left[\frac{1 + x^2}{(1 - xy)^2} \right]\left[\frac{1}{1 + x^2} \right] \] \[ = 0 \] Now, to find relationship between \(u\) and \(v\) \[ u = \frac{x + y}{1 - xy} \] \[ v = \tan^{-1}{x} + \tan^{-1}{y} \] \[ v = \tan^{-1}{\frac{x + y}{1 - xy}} \] \[ \left[\text{As, } \tan^{-1}{A} + \tan^{-1}{B} = \tan^{-1}{\left(\frac{A + B}{1 - AB} \right)} \right] \] \[ \Rightarrow \boxed{v = tan^{-1}{u}} \]
Sample Problem 9 :
Show that functions \[ \begin{align*} u &= x + y + z \\ v &= x^2 + y^2 + z^2 - 2xy -2yz - 2zx \\ w &= x^3 + y^3 + z^3 - 3xy^2 \end{align*} \] are functionally related. Find the relation.
Solution :
\[J\left(\frac{u,v,w}{x,y,z} \right) \] \[ = \Large \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{vmatrix} \] \[ = \begin{vmatrix} 1 & 1 & 1 \\ 2x - 2y - 2z & 2y - 2x - 2z & 2z - 2y - 2x \\ 3x^2 - 3yz & 3y^2 - 3xz & 3z^2 - 3xy \end{vmatrix} \] \[ = 1[(2y - 2x - 2z)(3z^2 - 3xy) - (2z - 2y - 2x)(3y^2 - 3zx)] \] \[ - 1[(2x - 2y - 2z)(3z^2- 3xy) - (2z - 2y - 2x)(3x^2 - 3yz)] \] \[ + 1[(2x - 2y - 2z)(3y^2 - 3zx) - (2y - 2x - 2z)(3x^2 - 3yz)] \] \[ = 0 \Rightarrow \text{ u and v are functionally related } \] \[ \text{Now, } w = x^3 + y^3 + z^3 - 3xyz \] \[ = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \] \[ = u[(v + 2xy + 2yz + 2zx) - xy - yz - zx] \] \[ = u(v + xy + yz + zx) \] \[ = \frac{4}{4} \times u(v + xy + yz + zx) \] \[ = \frac{u}{4}(4v + 4xy + 4yz + 4zx) \] \[ = \frac{u}{4}[4v + (2xy + 2yz + 2zx) - (-2xy - 2yz - 2zx)] \] \[ = \frac{u}{4}[4v + (x^2 + y^2 + z^2 + 2xy + 2yz + 2zx) - (x^2 + y^2 + z^2 - 2xy - 2yz - 2zx)] \] \[ = \frac{u}{4}[4v + (x + y + z)^2 - v] \] \[ = \frac{u}{4}(3v + u^2) \] \[ \Rightarrow \boxed{4w = u^3 + 3uv} \]
