A) \[{{x}^{3}}+qx+r=0\]
B) \[{{x}^{3}}-px-r=0\]
C) \[{{x}^{3}}+qx-r=0\]
D) None of these
Correct Answer: C
Solution :
Since, \[\alpha +i\beta \] is a root of the equation \[{{x}^{3}}+qx+r=0,\]\[\alpha -i\beta \] is also its roots. Let the third root be \[\gamma ,\] so that \[\alpha +i\beta +(\alpha -i\beta )+\gamma =0\] \[\Rightarrow \]\[{{x}^{3}}+qx-r=0\] Also, \[\gamma \] is the root of the given equation, therefore \[2\alpha \] is one of the roots of the equation\[{{(-x)}^{3}}+q(-x)+r=0\] \[\Rightarrow \]\[{{x}^{3}}+qx-r=0\]
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