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\a

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $\alpha^2, \beta^2, \gamma^2$. The sum of the new roots is $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (2)^2 - 2(10) = 4 - 20 = -16$. The sum of the products of the new roots taken two at a time is $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) = (10)^2 - 2(8)(2) = 100 - 32 = 68$. The product of the new roots is $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (8)^2 = 64$. The required cubic equation is $x^3 - (	ext{sum of roots})x^2 + (	ext{sum of product of roots taken two at a time})x - (	ext{product of roots}) = 0$. Substituting the values: $x^3 - (-16)x^2 + 68x - 64 = 0$,which simplifies to $x^3+16x^2+68x-64=0$.

Column-$I$	Column-$II$
$A$. $\alpha+\beta+\gamma$	$(1)$ $-\frac{43}{4}$
$B$. $\alpha^2+\beta^2+\gamma^2$	$(2)$ $-\frac{7}{8}$
$C$. $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$	$(3)$ $86$
$D$. $\frac{\alpha}{\beta\gamma}+\frac{\beta}{\gamma\alpha}+\frac{\gamma}{\alpha\beta}$	$(4)$ $0$
	$(5)$ $10$

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

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