A) \[0\]
B) \[3\,xyz\]
C) \[x+y+z\]
D) None of these
Correct Answer: B
Solution :
[b] \[D=\left| \begin{matrix} {{x}^{2}} & {{x}^{2}}-{{(y-z)}^{2}} & yz \\ {{y}^{2}} & {{y}^{2}}-{{(z-x)}^{2}} & zx \\ {{z}^{2}} & {{z}^{2}}-{{(x-y)}^{2}} & xy \\ \end{matrix} \right|\] \[=\left| \begin{matrix} {{x}^{2}} & -({{x}^{2}}+{{y}^{2}}+{{z}^{2}}) & yz \\ {{y}^{2}} & -({{x}^{2}}+{{y}^{2}}+{{z}^{2}}) & zx \\ {{z}^{2}} & -({{x}^{2}}+{{y}^{2}}+{{z}^{2}}) & xy \\ \end{matrix} \right|\] (Operating \[{{C}_{2}}\to {{C}_{2}}-2{{C}_{1}}-2{{C}_{3}}\]) \[=-({{x}^{2}}+{{y}^{2}}+{{z}^{2}})\left| \begin{matrix} {{x}^{2}} & 1 & yz \\ {{y}^{2}} & 1 & zx \\ {{z}^{2}} & 1 & xy \\ \end{matrix} \right|\] \[=-\frac{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}{xyz}\left| \begin{matrix} {{x}^{3}} & x & xyz \\ {{y}^{3}} & y & xyz \\ {{z}^{3}} & x & xyz \\ \end{matrix} \right|\] (Multiplying \[{{R}_{1}},{{R}_{2}}\] and \[{{R}_{3}}\] by x, y and z, respectively) \[=-({{x}^{2}}+{{y}^{2}}+{{z}^{2}})\left| \begin{matrix} {{x}^{3}} & x & 1 \\ {{y}^{3}} & y & 1 \\ {{z}^{3}} & x & 1 \\ \end{matrix} \right|\] \[=({{x}^{2}}+{{y}^{2}}+{{z}^{2}})\left| \begin{matrix} 1 & x & {{x}^{3}} \\ 1 & y & {{y}^{3}} \\ 1 & x & {{z}^{3}} \\ \end{matrix} \right|\] \[=(x-y)\,(y-z)\,(z-x)\,(x+y+z)\,({{x}^{2}}+{{y}^{2}}+{{z}^{2}})\] Given that\[~D=0\]. \[\Rightarrow \,\,\,\,\,\,x+y+z=0\] (as \[x\ne y\ne z\]) \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz\]
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