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If $\alpha,\beta $ and $ \gamma$ are the roots of the equation $x^3+3x^2-8x+1=0$, then an equation whose roots are $\alpha +1 , \beta +1 , \gamma +1 $ is given by

  1. $y^3-11y+11=0$
  2. $y^3-11y-11=0$
  3. $y^3+13y+13=0$
  4. $y^3+6y^2+y-3=0$
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We know that if  α , β  and  γ  are the roots of the equation ax3 + bx2 +cx +d

Then the product of the roots =  α  β  γ  = - (d/a)

and sum of the product of the roots taking two at a time is ( α  β  +  β  γ  + α  γ ) = c/a

and the sum of the roots ( α  + β  + γ ) = - (b/a)

So, in the given problem  α  β  γ  = -1  ( α  β  +  β  γ  + γ  α ) = -8 and ( α  + β  + γ ) = -3

Now, ( α +1) ( β +1)( γ +1) =  α  β  γ +( α  β  +  β  γ  +  α  γ ) + ( α + β + γ ) + 1 = (-1) + (-8) + (-3) + 1 = -11

So, the equation whose roots are ( α +1) , ( β +1) , ( γ +1) will be of the form  y3 + by2 + cy +11

which matches with option A

So, Option A will be the answer.  

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