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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

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B

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C

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D

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Solution

The correct option is C


x3 + 2x2 + 3x + 1 = 0

x3 + 1 = -(2 x2) + 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(-x3 -1)]

Putting x3 = y in this equation

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

y3 - 7y2 +12y + 1 = 0 {its root 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


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