The cubic equation whose roots are the square | vedclass

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Question

(A) Let $\alpha, \beta, \gamma$ be the roots of $x^3-2x^2+10x-8=0$.From Vieta's formulas:$\alpha+\beta+\gamma = 2$$\alpha\beta+\beta\gamma+\gamma\alph

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$x^3+16x^2+68x-64=0$

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(A) Let $\alpha, \beta, \gamma$ be the roots of $x^3-2x^2+10x-8=0$.From Vieta's formulas:$\alpha+\beta+\gamma = 2$$\alpha\beta+\beta\gamma+\gamma\alpha = 10$$\alpha\beta\gamma = 8$We want the equation with roots $\alpha^2, \beta^2, \gamma^2$.Sum of new roots: $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (2)^2 - 2(10) = 4 - 20 = -16$.Sum of roots taken two at a time: $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) = (10)^2 - 2(8)(2) = 100 - 32 = 68$.Product of new roots: $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (8)^2 = 64$.The required cubic equation is $x^3 - (	ext{sum of roots})x^2 + (	ext{sum of roots taken two at a time})x - (	ext{product of roots}) = 0$.Substituting the values: $x^3 - (-16)x^2 + 68x - 64 = 0$.Thus,$x^3+16x^2+68x-64=0$.

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

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