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Question

lf $$\alpha,\beta,\gamma$$ are the roots of the equation $$x^{3}-x+2=0$$, then the equation whose roots are $$\displaystyle \alpha \beta +\cfrac { 1 }{ \gamma  }, \displaystyle\beta\gamma+\frac{1}{\alpha},\gamma\alpha+\frac{1}{\beta}$$, is:


A
2y3+y2+1=0
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B
2y3y2+1=0
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C
y3+y2+1=0
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D
2y3+y21=0
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Solution

The correct option is B $$2y^{3}+y^{2}-1=0$$
As $$\alpha ,\beta ,\gamma $$ are roots of $${ x }^{ 3 }-x+2=0$$
We get $${ s }_{ 3 }=\alpha \beta \gamma =-2$$  ..(1)
Now $$y=\alpha \beta +\cfrac { 1 }{ \gamma  } \Rightarrow y=\cfrac { \alpha \beta \gamma +1 }{ \gamma  } $$ 
Using (1)
$$y=\cfrac { -2+1 }{ \gamma  } \Rightarrow y=-\cfrac { 1 }{ x } \Rightarrow x=-\cfrac { 1 }{ y } $$
Replacing $$x\rightarrow -\cfrac { 1 }{ y } $$ in $${ x }^{ 3 }-x+2=0$$ , we get
$${ \left( -\cfrac { 1 }{ y }  \right)  }^{ 3 }-\left( -\cfrac { 1 }{ y }  \right) +2=0\Rightarrow -1+{ y }^{ 2 }+2{ y }^{ 3 }=0\\ \Rightarrow 2{ y }^{ 3 }+{ y }^{ 2 }-1=0$$

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