A) \[{{p}^{3}}+{{q}^{2}}-q(3p+1)=0\]
B) \[{{p}^{3}}+{{q}^{2}}+q(1+3p)=0\]
C) \[{{p}^{3}}+{{q}^{2}}+q(3p-1)=0\]
D) \[{{p}^{3}}+{{q}^{2}}+q(1-3p)=0\]
Correct Answer: D
Solution :
Let root of the given equation \[{{x}^{2}}+px+q=0\] are \[\alpha \] and \[{{\alpha }^{2}}\]. Now, \[\alpha .{{\alpha }^{2}}={{\alpha }^{3}}=q,\] \[\alpha +{{\alpha }^{2}}=-p\] Cubing both sides, \[{{\alpha }^{3}}+{{({{\alpha }^{2}})}^{3}}+3\alpha \,.\,{{\alpha }^{2}}(\alpha +{{\alpha }^{2}})=-{{p}^{3}}\] \[q+{{q}^{2}}+3q(-p)=-{{p}^{3}}\] \[{{p}^{3}}+{{q}^{2}}+q(1-3p)=0\].
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