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 $γ^{3}$ .

$x^{3} - 5 x^{2} + 5 x + 2 = 0$

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$x^{3} - x^{2} + 15 x + 2 = 0$

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$x^{3} - 7 x^{2} + 12 x + 1 = 0$

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$x^{3} - 2 x^{2} + 10 x + 1 = 0$

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Solution

The correct option is C
$x^{3} - 7 x^{2} + 12 x + 1 = 0$

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

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

Cubing on both sides

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

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

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

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

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

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

$\Rightarrow y^{3} - 7 y^{2} + 12 y + 1 = 0$
Its roots will be $α^{3}, β^{3}, γ^{3}$

In terms of $x$ , we have
$x^{3} - 7 x^{2} + 12 x + 1 = 0$

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