The cubic equation whose roots are the squares of the roots of the equation $12 x^3-20 x^2+x+3=0$ is

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Aicharya · Accepted Answer

Let $\alpha, \beta, \gamma$ be the roots of the equation $\begin{aligned} & 12 x^3-20 x^2+x+3=0 \ & \because \alpha+\beta+\gamma=\frac{20}{12} \ & \alpha \beta+\beta \gamma+\alpha \gamma=\frac{1}{12}, \alpha \beta \gamma=\frac{-3}{12} \end{aligned}$ Now, $(\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha \beta+\beta \gamma+\alpha \gamma)$ $\begin{aligned} & \begin{array}{l} \Rightarrow \alpha^2+\beta^2+\gamma^2=\frac{376}{144} \ 	ext { and, }(\alpha \beta+\beta \gamma+\alpha \gamma)^2=(\alpha \beta)^2+(\beta \gamma)^2+(\alpha \gamma)^2 \ +2 \alpha \beta \gamma(\alpha+\beta+\gamma) \ \Rightarrow(\alpha \beta)^2+(\beta \gamma)^2+(\alpha \gamma)^2=\frac{121}{144} \end{array} \end{aligned}$ Now, required equation is $\begin{aligned} & x^3-\left(\alpha^2+\beta^2+\gamma^2ight) x^2+\left(\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2ight) x \ & \Rightarrow 144 x^3-376 x^2+121 x-9=0 . \end{aligned}$

The cubic equation whose roots are the squares of the roots of the equation $12 x^3-20 x^2+x+3=0$ is

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