If the roots of the equation x^3-6x^2+11x-6=0 | vedclass

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(B) Given the equation $x^3-6x^2+11x-6=0$ with roots $\alpha, \beta, \gamma$. Let $y = x^2$,so $x = \sqrt{y}$. Substituting into the original equation

Vedclass · Accepted Answer

$x^3-14x^2+49x-36=0$

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(B) Given the equation $x^3-6x^2+11x-6=0$ with roots $\alpha, \beta, \gamma$. Let $y = x^2$,so $x = \sqrt{y}$. Substituting into the original equation: $(\sqrt{y})^3 - 6(\sqrt{y})^2 + 11\sqrt{y} - 6 = 0$. $y\sqrt{y} - 6y + 11\sqrt{y} - 6 = 0$. Rearranging terms: $\sqrt{y}(y+11) = 6(y+1)$. Squaring both sides: $y(y+11)^2 = 36(y+1)^2$. $y(y^2 + 22y + 121) = 36(y^2 + 2y + 1)$. $y^3 + 22y^2 + 121y = 36y^2 + 72y + 36$. $y^3 - 14y^2 + 49y - 36 = 0$. Replacing $y$ with $x$,the required equation is $x^3-14x^2+49x-36=0$.

If the roots of the equation $x^3-6x^2+11x-6=0$ are $\alpha, \beta$ and $\gamma$,then the equation whose roots are $\alpha^2, \beta^2, \gamma^2$ is:

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