Question

If α, β, γ are roots of equation x3 +px + q = 0

Answer 1

α³+β³+γ³=3q−p(α+β+γ)
=3q−p(0)=3q
(α+β)(β+γ)(γ+α)=(α+β+γ)³−(α³+β³+γ³)³/3
=0³−p³−3q³/3
αβγ=−q, and also αβ+βγ+γα=p

Answer 2

$\sf If \: \alpha, \beta , and \: \gamma \: are \: roots \: of \: f(x)$

$f(x) = \sf {x}^{3} + px + q = 0$

Thus,

The Relations between the roots and the coefficients are:

$\alpha + \beta + \gamma = 0$

$\alpha \beta + \alpha \gamma + \beta \gamma = p$

$\alpha \beta \gamma = - q$

And, the values of the three roots are:

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$\\ \sf \alpha = \sqrt[3]{ - \frac{q}{2} + \sqrt{(\frac{q}{2} ) {}^{2} + {( \frac{p}{3} ) {}^{3} }} } + \: \sf \sqrt[3]{ - \frac{q}{2} - \sqrt{(\frac{q}{2} ) {}^{2} + {( \frac{p}{3} ) {}^{3} }} }$

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$\sf\beta = \sqrt[3]{ \frac{q}{16} + \sqrt{ (\frac{q}{16} )^{2} + { (\frac{p}{12} ) {}^{3} }} }+ \sqrt[3]{ \frac{q}{16} - \sqrt{ (\frac{q}{16} )^{2} + { (\frac{p}{12} ) {}^{3} }} } \\ \sf + \it{i} \sf\frac{ \sqrt{3} }{2} (\sqrt[3]{ \frac{q}{16} + \sqrt{ (\frac{q}{16} )^{2} + { (\frac{p}{12} ) {}^{3} }} } - \sqrt[3]{ \frac{q}{16} - \sqrt{ (\frac{q}{16} )^{2} + { (\frac{p}{12} ) {}^{3} }} })$

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$\sf\gamma= \sqrt[3]{ \frac{q}{16} + \sqrt{ (\frac{q}{16} )^{2} + { (\frac{p}{12} ) {}^{3} }} }+ \sf \sqrt[3]{ \frac{q}{16} - \sqrt{ (\frac{q}{16} )^{2} + { (\frac{p}{12} ) {}^{3} }} } \\ \sf - \it{i} \sf\frac{ \sqrt{3} }{2} (\sqrt[3]{ \frac{q}{16} + \sqrt{ (\frac{q}{16} )^{2} + { (\frac{p}{12} ) {}^{3} }} } - \sqrt[3]{ \frac{q}{16} - \sqrt{ (\frac{q}{16} )^{2} + { (\frac{p}{12} ) {}^{3} }} })$

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where,

$\it{i} = \sf \sqrt{-1}$

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