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If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3-4x^2+x+6=0$, find the equation whose roots are

1) $1/\alpha$, $1/\beta$, $1/\gamma$

2) $\alpha^2$, $\beta^2$, $\gamma^2$

How to do these kind of questions. I am stuck. I know,

$\alpha+\beta+\gamma=-b/a$

$\alpha\beta+\beta \gamma+\alpha \gamma=c/a$

$\alpha \beta\gamma=-d/a$

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  • $\begingroup$ These are not "quadratic roots". You should fix the title. $\endgroup$ – Yves Daoust Jun 20 '17 at 15:34
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For the first one, set $x=1/y\implies$ $$\left(\dfrac1y\right)^3-4\left(\dfrac1y\right)^2+\left(\dfrac1y\right)+6=0$$

Multiply by $y^3$

For the second, $x^2=y,$

$$x(x^2+1)=4x^2-6\implies x(y+1)=4y-6$$

Square both sides and replace $x^2$ with $y$

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  • $\begingroup$ Why did you put x=1/y @labbhattacharjee $\endgroup$ – onlymaths Jun 20 '17 at 15:29
  • $\begingroup$ They are just representing $x$ in a form more useful to answer the question. $\endgroup$ – fosho Jun 20 '17 at 15:32
  • $\begingroup$ @onlymaths, You can see the expected pattern of the roots $\endgroup$ – lab bhattacharjee Jun 20 '17 at 15:56
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This equation can be factored as $(x + 1)(x - 2) (x -3)$. Let $\alpha = -1$, $\beta = 2$, and $\gamma = 3$. To find the equation with inverses of all of these, simply find the inverses and put them all together, such that you have $(x + 1)(x - 1/2)(x - 1/3)$. The same goes for the squares $(x - 1)(x - 4) (x-9)$.

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for 1/alpha +1/beta +1/gamma just do sum of roots(2 at a time)/product of roots. For e.g. alphabeta+alphagamma+bettagamma/alphabetagamma

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  • 1
    $\begingroup$ Please explain why. Also, take a look at using TeX to display formulae. $\endgroup$ – Rolazaro Azeveires Sep 9 '18 at 13:05
  • $\begingroup$ @Genius Your username doesn't come off as polite. Would you mind changing it? $\endgroup$ – Toby Mak Sep 9 '18 at 13:34

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