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Question Number 78814 by M±th+et£s last updated on 20/Jan/20

Commented by mind is power last updated on 21/Jan/20

$$2ndWay$$$${\alpha }^2{\beta }^2{\gamma }^2{\delta }^2+{\alpha }^2{\beta }^2+{\alpha }^2{\gamma }^2+{\alpha }^2{\delta }^2+{\beta }^2{\gamma }^2+{\beta }^2{\delta }^2+{\gamma }^2{\delta }^2+{\alpha }^2{\beta }^2{\gamma }^2+{\alpha }^2{\beta }^2{\delta }^2+{\beta }^2{\gamma }^2{\delta }^2+{\alpha }^2{\gamma }^2{\delta }^2$$$$+1+{\alpha }^2+{\beta }^2+{\delta }^2+{\gamma }^2$$$${\alpha }^2+{\beta }^2+{\delta }^2+{\gamma }^2={(\alpha +\beta +\delta +\gamma )}^2-2(\alpha \beta +\beta \delta +\beta \gamma +\alpha \delta +\alpha \gamma +\delta \gamma )$$$$={(-b)}^2-2c=b^2-2c$$$${\alpha }^2{\beta }^2{\gamma }^2{\delta }^2={\left(\alpha \beta \gamma \delta \right)}^2=e^2$$$$({\alpha }^2{\beta }^2+{\alpha }^2{\delta }^2+{\alpha }^2{\gamma }^2+{\beta }^2{\delta }^2+{\beta }^2{\gamma }^2+{\delta }^2{\gamma }^2)={(\alpha \beta +\alpha \delta +\alpha \gamma +\beta \delta +\beta \gamma +\delta \gamma )}^2$$$$S=\alpha \beta +\beta \gamma +\beta \delta +\delta \gamma +\delta \alpha +\gamma \alpha$$$$-2{\alpha }^2\beta \delta -2{\alpha }^2\beta \gamma -2{\alpha }^2\delta \gamma -2{\beta }^2(\alpha \gamma +\gamma \delta +\alpha \delta )-2{\gamma }^2(\alpha \beta +\alpha \delta +\delta \beta )$$$$-2{\delta }^2(\alpha \beta +\alpha \gamma +\gamma \beta )$$$$=c^2-2{\alpha }^2(S-\alpha (\beta +\gamma +\delta ))-2{\beta }^2(S-\beta (\alpha +\gamma +\delta ))-2{\delta }^2(S-\delta (\alpha +\beta +\gamma ))$$$$=-2{\gamma }^2(S-\gamma (\alpha +\beta +\delta ))$$$$=-2S({\alpha }^2+{\beta }^2+{\delta }^2+{\gamma }^2)+2{\alpha }^3(\beta +\gamma +\delta )+2{\beta }^3(\alpha +\gamma +\delta )+2{\delta }^2(\alpha +\beta +\gamma )$$$$+2{\gamma }^3(\alpha +\beta +\delta )$$$$U=\alpha +\beta +\delta +\gamma$$$$=-2S({\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2)+2U({\alpha }^3+{\beta }^3+{\delta }^3+{\gamma }^3)-2{\alpha }^4-2{\beta }^4-2{\gamma }^4-2{\delta }^4$$$${\alpha }^3+{\beta }^3+{\gamma }^3+{\delta }^3$$$$useNewtoonIdentity$$$$TooBeecontinuednottheBestWaytogoforbutWorcks$$

Answered by mind is power last updated on 20/Jan/20

$$({\mathrm{x}}^2+1)=(\mathrm{x}+\mathrm{i})(\mathrm{x}-\mathrm{i})$$$$\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}=(\mathrm{x}-\alpha )(\mathrm{x}-\beta )(\mathrm{x}-\gamma )(\mathrm{x}-\delta )$$$$({\alpha }^2+1)({\beta }^2+1)({\gamma }^2+1)({\delta }^2+1)$$$$=(\alpha -\mathrm{i})(\beta -\mathrm{i})(\gamma -\mathrm{i})(\delta -\mathrm{i})(\alpha +\mathrm{i})(\beta +\mathrm{i})(\gamma +\mathrm{i})(\delta +\mathrm{i})$$$$\mathrm{p}(-\mathrm{i})=(-\mathrm{i}-\alpha )(-\mathrm{i}-\beta )(-\mathrm{i}-\gamma )(-\mathrm{i}-\delta )$$$$=(\mathrm{i}+\alpha )(\mathrm{i}+\beta )(\mathrm{i}+\gamma )(\mathrm{i}+\delta )$$$$\mathrm{p}\left(\mathrm{i}\right)=(\mathrm{i}-\alpha )(\mathrm{i}-\gamma )(\mathrm{i}-\delta )(\mathrm{i}-\beta )=(\alpha -\mathrm{i})(\beta -\mathrm{i})(\delta -\mathrm{i})(\gamma -\mathrm{i})$$$$\iff ({\alpha }^2+1)({\beta }^2+1)({\gamma }^2+1)({\delta }^2+1)=\mathrm{p}\left(\mathrm{i}\right).\mathrm{p}(-\mathrm{i})$$$$=(1-\mathrm{ia}-\mathrm{b}+\mathrm{ci}+\mathrm{d})(1+\mathrm{ia}-\mathrm{b}-\mathrm{ci}+\mathrm{d})$$$$=(1-\mathrm{b}+\mathrm{d}+\mathrm{i}(\mathrm{d}-\mathrm{a}))(1-\mathrm{b}+\mathrm{d}-\mathrm{i}(\mathrm{d}-\mathrm{a}))$$$$\mathrm{if}\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e}\mathrm{are}\mathrm{reel}$$$$={(1-\mathrm{b}+\mathrm{d})}^2+{(\mathrm{d}-\mathrm{a})}^2$$$$\mathrm{other}\mathrm{case}=(1-\mathrm{b}+\mathrm{d}+\mathrm{i}(\mathrm{d}-\mathrm{a}))(1-\mathrm{b}+\mathrm{d}-\mathrm{i}(\mathrm{d}-\mathrm{a}))$$

Commented by M±th+et£s last updated on 20/Jan/20

$$nicejob.thoughthereisasmallerror$$$$inthefourthlinefrombelowwhen$$$$youwanttofactorizebyi,its:\mathrm{i}(\mathrm{c}-\mathrm{a})and-\mathrm{i}(\mathrm{c}-\mathrm{a}).$$

Commented by mind is power last updated on 20/Jan/20

$$\mathrm{yeah}$$

Commented by jagoll last updated on 21/Jan/20

$$\mathrm{mr}\mathrm{Mind}\mathrm{is}\mathrm{power}$$$$\mathrm{if}\mathrm{we}\mathrm{work}\mathrm{using}{\mathrm{Vieta}}^{\prime }\mathrm{s}\mathrm{rule},$$$$\mathrm{given}\mathrm{the}\mathrm{same}\mathrm{result}?$$

Commented by mind is power last updated on 21/Jan/20

$$\mathrm{yeah}$$

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