The cubic equation whose roots are thrice 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 D $x^{3} + 6 x^{2} - 36 x + 27 = 0$
Given cubic equation is $x^{3} + 2 x^{2} - 4 x + 1 = 0$
Replacing $x \to \frac{x}{3}$ in $x^{3} + 2 x^{2} - 4 x + 1 = 0$ , we get
${(\frac{x}{3})}^{3} + 2 {(\frac{x}{3})}^{2} - 4 (\frac{x}{3}) + 1 = 0 \Rightarrow \frac{x^{3}}{27} + \frac{2 x^{2}}{9} - \frac{4 x}{3} + 1 = 0$
$\Rightarrow x^{3} + 6 x^{2} - 36 x + 27 = 0$

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