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Question Number 192186 by Tawa11 last updated on 11/May/23

$$\mathrm{If}\alpha ,\beta \mathrm{and}\gamma \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}{\mathrm{x}}^3+\mathrm{px}+\mathrm{q}=0,\mathrm{find}\mathrm{\Sigma }{\alpha }^4.$$

Answered by BaliramKumar last updated on 10/May/23

$${2\mathrm{p}}^2$$

Commented by Tawa11 last updated on 10/May/23

$$\mathrm{Please}\mathrm{workings}\mathrm{sir}?$$

Answered by manxsol last updated on 11/May/23

$$\alpha +\beta +\gamma =0$$$$\alpha \beta +\alpha \gamma +\beta \gamma =p$$$$\alpha \beta \gamma =-q$$$${\alpha }^4+p{\alpha }^2+q\alpha =0$$$${\beta }^4+p{\beta }^2+q\beta =0$$$${\gamma }^4+p{\gamma }^2+q\gamma =0$$$${\alpha }^4+{\beta }^4+{\gamma }^4+p({\alpha }^2+{\beta }^2+{\gamma }^2)+q(\alpha +\beta +\gamma =0$$$${(\alpha +\beta +\gamma )}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\alpha \gamma +\beta \gamma )=0$$$${\left(0\right)}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2\left(p\right)$$$$-2p={\alpha }^2+{\beta }^2+{\gamma }^2$$$${\alpha }^4+{\beta }^4+{\gamma }^4+p(-2p)+q\left(0\right)=0$$$${\alpha }^4+{\beta }^4+{\gamma }^4=2p^2$$

Commented by Tawa11 last updated on 11/May/23

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by BaliramKumar last updated on 11/May/23

$$\alpha +\beta +\gamma =0$$$$\alpha \beta +\beta \gamma +\gamma \alpha =\mathrm{p}$$$$\alpha \beta \gamma =-\mathrm{q}$$$$\mathrm{\Sigma }{\alpha }^4={\alpha }^4+{\beta }^4+{\gamma }^4={\left({\alpha }^2\right)}^2+{\left({\beta }^2\right)}^2+{\left({\gamma }^2\right)}^2$$$$\Rightarrow {({\alpha }^2+{\beta }^2+{\gamma }^2)}^2-2({\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2)$$$$\Rightarrow {\{{(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )\}}^2$$$$-2\{{(\alpha \beta +\beta \gamma +\gamma \alpha )}^2-2(\alpha {\beta }^2\gamma +\alpha \beta {\gamma }^2+{\alpha }^2\beta \gamma )\}$$$$\Rightarrow {\{{(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )\}}^2$$$$-2\{{(\alpha \beta +\beta \gamma +\gamma \alpha )}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )\}$$$$\Rightarrow {\{{\left(0\right)}^2-2\left(\mathrm{p}\right)\}}^2-2\{{\left(\mathrm{p}\right)}^2-2(-\mathrm{q})\left(0\right)\}$$$$\Rightarrow {\{-2\mathrm{p}\}}^2-2\{{\mathrm{p}}^2-0\}$$$$\Rightarrow {4\mathrm{p}}^2-{2\mathrm{p}}^2$$$$\Rightarrow \overline{)\begin{array}{c}{2\mathrm{p}}^2\end{array}}$$$$$$$$$$$$$$$$$$$$$$

Commented by Tawa11 last updated on 11/May/23

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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