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The cubic equation $x^3−2x^2−3x+4=0$ has the roots $α$ , $β$ and $γ$. Using the substitution method how do I find an equation with the roots $ \frac{1}{(α+β)^2}$, $ \frac{1}{(α+γ)^2}$ , $ \frac{1}{(β+γ)^2}$ .

I tried doing it with $\sigma_1$ , $\sigma_2$ and $\sigma_3$ where
$\sigma_1$ is ($\alpha + \beta + \gamma$) ,
$\sigma_2$ is ($\alpha\beta + \alpha\gamma + \beta\gamma)$ and
$\sigma_3$ is ($\alpha\beta\gamma$)

But it is a lengthy method and I guess it will be easier to use the substitutuion method .

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Hint $$\alpha + \beta +\gamma =2 \implies \alpha + \beta =2-\gamma $$

So all you need to do is, substitute $$y= \frac1{(2-x)^2}$$

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  • $\begingroup$ So the answer is , 2(y^3/2) - 4(y^1/2) + 5y - 1 = 0 ? $\endgroup$ – TheOppositeGuy Feb 17 '18 at 5:42
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    $\begingroup$ I didn't check the answer, but if you substituted $x=2-\frac1{\sqrt y} $ and then proceeded by simplifying it, you are correct. $\endgroup$ – Jaideep Khare Feb 17 '18 at 5:55
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    $\begingroup$ @The But do note that, if you have to apply transformation of roots, then the final answer MUST also be a polynomial of same degree, so you need to simplify further, until you finally get a cubic in $y$. For doing that, collect $y^{3/2}$ and $y^{1/2}$ terms on LHS, rest on RHS, and square both the side, then simplify. $\endgroup$ – Jaideep Khare Feb 17 '18 at 6:04

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