The cubic equation whose roots are thrice to | vedclass

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(D) Let the roots of the given equation $x^3+2x^2-4x+1=0$ be $\alpha, \beta, \gamma$. We want to find the equation whose roots are $3\alpha, 3\beta, 3

Vedclass · Accepted Answer

$x^3+6x^2-36x+27=0$

Vedclass · Answer

(D) Let the roots of the given equation $x^3+2x^2-4x+1=0$ be $\alpha, \beta, \gamma$. We want to find the equation whose roots are $3\alpha, 3\beta, 3\gamma$. Let $y = 3x$,which implies $x = \frac{y}{3}$. Substituting $x = \frac{y}{3}$ into the original equation: $(\frac{y}{3})^3 + 2(\frac{y}{3})^2 - 4(\frac{y}{3}) + 1 = 0$ $\frac{y^3}{27} + \frac{2y^2}{9} - \frac{4y}{3} + 1 = 0$ Multiplying the entire equation by $27$: $y^3 + 6y^2 - 36y + 27 = 0$ Replacing $y$ with $x$,the required equation is $x^3+6x^2-36x+27=0$.

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

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