The cubic equation whose roots are the squares of the roots of $x^3-2 x^2+10 x-8=0$, is

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Aicharya · Accepted Answer

Let, $\alpha, \beta$ and $\gamma$ are the roots of $ \begin{aligned} & x^3-2 x^2+10 x-8 & =0 \ 	herefore & \alpha+\beta+\gamma & =2, \end{aligned} $ $ \alpha \beta+\beta \gamma+\gamma \alpha=10 $ and $\alpha \beta \gamma=8$ Now, $\alpha^2+\beta^2+\gamma^2$ $ \begin{aligned} & =(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha) \ & =(2)^2-2(10)=-16 \end{aligned} $ $ \begin{aligned} \alpha^2 \beta^2+ & \beta^2 \gamma^2+\gamma^2 \alpha^2 \ & =(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2(\alpha+\beta+\gamma)(\alpha \beta \gamma) \ & =(10)^2-2(2)(8) \ & =100-32=68 \end{aligned} $ and $ \alpha^2 \beta^2 \gamma^2=(8)^2=64 $ $	herefore$ Required cubic 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+\gamma^2 \alpha^2ight) x \ & -\alpha^2 \beta^2 \gamma^2=0 \ & 	herefore x^3-(-16) x^2+(68) x-64=0 \ & \Rightarrow \quad x^2+16 x^2+68 x-64=0 \end{aligned} $

The cubic equation whose roots are the squares of the roots of $x^3-2 x^2+10 x-8=0$, is

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