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Question

If α,β and γ are the roots of the equation x3+2x2+3x+1=0. Find the equation whose roots are 1α3,1β3,1γ3


A

x3+12x27x+1=0

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B

x3x2+8x+2=0

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C

x3+15x28x+1=0

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D

x3x2+12x+1=0

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Solution

The correct option is A

x3+12x27x+1=0


x3+2x2+3x+1=0

x3+1=(2x2+3x)...(1)

Cubing on both sides

(x3+1)3=(2x2+3x)3

x9+3x6+3x3+1=[8x6+27x3+18x3(2x2+3x)]

x3 + 1 = -(2x2 + 3x) from equation (1).

x9+3x6+3x3+1=[8x6+27x3+18x3(x31)]

Putting x3 = y in this equation

y3 + 3y2 +3y + 1 = [8y2+27y+18y(y1)]

y37y2+12y+1=0
Its roots will be α3,β3,γ3 ...(2)

To get the equation whose roots are 1α3,1β3,1γ3, Change y to 1y in equation (2)

1y37y2+12y+1=0

Hence required equation in y is
y3+12y27y+1=0
Change the variable to x, we have
x3+12x27x+1=0


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