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Question

If α and β are the roots of the quadratic equation (x2)(x3)+(x3)(x+1)+(x+1)(x2)=0, then the value of 1(α+1)(β+1)+1(α2)(β2)+1(α3)(β3) is

A
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B
1
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C
0
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D
2
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Solution

The correct option is C 0
(x2)(x3)+(x3)(x+1)+(x+1)(x2)=0x25x+6+x22x3+x2x2=03x28x+1=0
The sum and product of the roots are,
α+β=83 and αβ=13
Now,
1(α+1)(β+1)+1(α2)(β2)+1(α3)(β3)=1αβ+(α+β)+1+1αβ2(α+β)+4+1αβ3(α+β)+9=141+34=0

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