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Question Number 69644 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19

Answered by MJS last updated on 26/Sep/19

$$\mathrm{let}$$$$x=\alpha$$$$y=\beta -\sqrt{\gamma }$$$$z=\beta +\sqrt{\gamma }$$$$\{\begin{array}{l}\alpha +2\beta -12=0\\ {\alpha }^2-2{\beta }^2+2c-12=0\\ {\alpha }^3+2{\beta }^3+6\beta \gamma -12=0\end{array}$$$$\Rightarrow$$$$\{\begin{array}{l}\alpha =-2\beta +12\\ \gamma =-3{\beta }^2+24\beta -66\\ {\beta }^3-12{\beta }^2+\frac{105}{2}\beta -\frac{143}{2}=0\end{array}$$$$$$$$x^4+y^4+z^4=\delta$$$$\Rightarrow$$$${\beta }^3-12{\beta }^2+\frac{105}{2}\beta -\frac{1227}{16}=-\frac{\delta }{384}$$$$\Rightarrow$$$$\delta =1992$$$$$$$$x^4+y^4+z^4=1992$$

Commented by mind is power last updated on 26/Sep/19

$$VerryNiceputyandZ={\beta }_{-}^{+}\sqrt{\gamma }$$$$$$

Answered by ajfour last updated on 26/Sep/19

$${(x+y+z)}^3=x^3+y^3+z^3+$$$$\mathrm{\Sigma }3xy(12-z)+6xyz$$$$\Rightarrow {\left(12\right)}^3=12-3xyz+36\mathrm{\Sigma }xy..\left(i\right)$$$${(x+y+z)}^2=\mathrm{\Sigma }{x}^2+2\mathrm{\Sigma }xy$$$$\Rightarrow \mathrm{\Sigma }xy=66$$$$nowfrom\left(i\right)$$$$3xyz=3p=12+36\times 66-{\left(12\right)}^3$$$$=12\{1+192-144\}$$$$=12\times 49$$$${\left(\mathrm{\Sigma }{x}^2\right)}^2=\mathrm{\Sigma }{x}^4+2\mathrm{\Sigma }{x}^2{y}^2$$$$=Q+2\{{\left(\mathrm{\Sigma }xy\right)}^2-2p\mathrm{\Sigma }x\}$$$$\Rightarrow Q=144-2\{{\left(66\right)}^2-8\times 12\times 49\}$$$$=144-24\{363-392\}$$$$=144+24\times 29=24\times 35$$$$=840.$$

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