A) 0
B) 3
C) 6
D) 12
Correct Answer: B
Solution :
\[\frac{1}{xyz}\left| \begin{matrix} {{x}^{4}}+x & {{x}^{3}}y & {{x}^{3}}z \\ x{{y}^{3}} & {{y}^{4}}+y & {{y}^{3}}z \\ x{{z}^{3}} & y{{z}^{3}} & {{z}^{4}}+z \\ \end{matrix} \right|=11\] \[\frac{xyz}{xyz}\left| \begin{matrix} {{x}^{3}}+1 & {{x}^{3}} & {{x}^{3}} \\ {{y}^{3}} & {{y}^{3}}+1 & {{y}^{3}} \\ {{z}^{3}} & {{z}^{3}} & {{z}^{3}}+1 \\ \end{matrix} \right|=11\] use\[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\] \[D=({{x}^{3}}+{{y}^{3}}+{{z}^{3}}+1)\]\[\left| \begin{matrix} 1 & 1 & 1 \\ {{y}^{3}} & {{y}^{3}}+1 & {{y}^{3}} \\ {{z}^{3}} & {{z}^{3}} & {{z}^{3}}+1 \\ \end{matrix} \right|=11\] Hence,\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=10\] \[(2,1,1),(1,2,1),(1,1,2)\]
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