If $α, β a n d γ$ are the roots of the equation $x^{3} + 2 x^{2} + 3 x + 1 = 0$ . Find the equation whose roots are $α^{3}$ , $β^{3}$ and $γ^{3}$ .

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Solution

The correct option is B

$x^{3}$ + 2 $x^{2}$ + 3x + 1 = 0

$x^{3} + 1 = - (2 x^{2}) + 3 x)$ ...................(1)

Cubing on both sides

${(x^{3} + 1)}^{3}$ = - ${(2 x^{2} + 3 x)}^{3}$

$x^{9}$ + 3 $x^{6}$ + 3 $x^{3}$ + 1 = -[8 $x^{6}$ +27 $x^{3}$ + 18 $x^{3}$ (2 $x^{2}$ + 3x))]

$x^{3}$ + 1 = -(2 $x^{2}$ + 3x from equation 1).

$x^{9}$ + 3 $x^{6}$ + 3 $x^{3}$ + 1 = -[8 $x^{6}$ + 27 $x^{3}$ + 18 $x^{3}$ (- $x^{3}$ -1)]

Putting $x^{3}$ = y in this equation

$y^{3}$ + 3 $y^{2}$ +3y + 1 = -[8 $y^{2}$ +27y + 18y (- y - 1)]

$y^{3}$ - 7 $y^{2}$ +12y + 1 = 0 {its root $α^{3}$ , $β^{3}$ , $γ^{3}$ }_____________(2)

Equation $x^{3}$ - $x^{2}$ + 12x + 1 = 0 has roots $α^{3}$ , $β^{3}$ , $γ^{3}$

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