The cubic equation whose roots are thrice to each of the roots of $x^{3} + 2 x^{2} - 4 x + 1 = 0$ is

x^{3} - 6 x^{2} + 36 x + 27 = 0

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x^{3} + 6 x^{2} + 36 x + 27 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{3} - 6 x^{2} - 36 x + 27 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{3} + 6 x^{2} - 36 x + 27 = 0

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Solution

The correct option is B $x^{3} + 6 x^{2} - 36 x + 27 = 0$
Given equation is $x^{3} + 2 x^{2} - 4 x + 1 = 0$
Let $α, β$ and $γ$ be the roots of the given equation
$∴ α + β + γ = - 2, α β + β γ + γ α = - 4$ and $α β γ = - 1$
Let the required cubic equation has the roots $3 α, 3 β$ and $3 γ$ .
$\Rightarrow 3 α + 3 β + 3 γ = - 6$ ,
$3 α \cdot 3 β + 3 β + 3 γ + 3 γ \cdot 3 α = 36$
and $3 α \cdot 3 β \cdot 3 γ = - 27$
$∴$ required equation is
$x^{3} - (- 6) x^{2} + (- 36) x - (- 27) = 0$
$\Rightarrow x^{3} + 6 x^{2} - 36 x + 27 = 0$

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