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if u, v,w are the roots of the equation (x-a)3+(x-b)3+(x-c)3=0 then find ∂(u,v,w)/∂(a,b,c) |
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Answer» Given that u, v, w are roots of the equation (x-a)3 + (x-b)3 + (x-c)3 = 0 ⇒ 3x3 - 3(a+b+c)x2 + 3(a2+b2+c2)x - (a3+b3+c3) = 0 ∵ u, v and w are roots of given equation. ∴ Sum of roots = u+v+w = -b/a \(=\frac{-(-3(a+b+c))}{3}\) = a+b+c ⇒ u + v + w - a - b - c = 0 ⇒ f(u, v, w) = u+v+w - (a+b+c) = 0 Now, \(\frac{\partial u}{\partial a}=1,\frac{\partial u}{\partial b}=1,\frac{\partial u}{\partial c}=1,\) \(\frac{\partial v}{\partial a}=1,\frac{\partial v}{\partial b}=1,\frac{\partial v}{\partial c}=1,\) \(\frac{\partial w}{\partial a}=1,\frac{\partial w}{\partial b}=1,\frac{\partial w}{\partial c}=1\) Now, \(\frac{\partial(u,v,w)}{\partial(a,b,c)}=\begin{bmatrix}\frac{\partial u}{\partial a}&\frac{\partial u}{\partial b}&\frac{\partial u}{\partial c}\\\frac{\partial v}{\partial a}&\frac{\partial v}{\partial b}&\frac{\partial v}{\partial c}\\\frac{\partial w}{\partial a}&\frac{\partial w}{\partial b}&\frac{\partial w}{\partial c}\end{bmatrix}\) \(=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}=0\) Hence, \(\frac{\partial(u,v,w)}{\partial(a,b,c)}=0\) |
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