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Question

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


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Solution

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 α3β3γ3}_____________(2)

Changing y to 1y in equation 2

1y37y212y + 1 = 0

y3 + 12y2 - 7y + 1 = 0 ____________(3)

If roots are {1α31β31γ3}

Let  1α3 = a, 1β3 = b, 1γ3 = c

a + b + c = -12

1β31γ31α3 = (a + b + c) - 2a = -12 - 2y

y = 12+p2

Putting the value of y in equation 3

-(12+p)38 + 12 (12+p)24 + 7 (12+p)2 + 1 = 0

(12+p)3 - 24(12+p)2 - 28(12 + p) - 8 = 0

p3 + 12p2 - 172p - 2072 = 0

Change the variable x in place of p

x3 + 12x2 - 172x - 2072

Constant term = -2072

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