A) \[2{{p}^{3}}-3pq+r=0\]
B) \[2{{p}^{3}}+3pq+r=0\]
C) \[2{{p}^{3}}-3pq-r=0\]
D) \[{{p}^{3}}+3pq-r=0\]
Correct Answer: A
Solution :
Let \[\alpha -\beta ,\,\alpha ,\,\alpha +\beta \] be the roots of the equation\[{{x}^{3}}+3p{{x}^{2}}+3qx+r=0\]. \[\therefore \] Sum of the roots \[=-\frac{b}{a}\] \[\Rightarrow \] \[\alpha -\beta +\alpha +\alpha +\beta =\frac{-3p}{1}\] \[\Rightarrow \] \[3\alpha =-3p\] \[\Rightarrow \] \[\alpha =-p\] \[\because \] \[\alpha \] satisfies the given equation. Put \[x=-p,\] we get \[-{{p}^{3}}+3{{p}^{3}}-3pq+r=0\] \[\Rightarrow \] \[2{{p}^{3}}-3pq+r=0\]
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