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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

let x=α y=β−(√γ) z=β+(√γ) { ((α+2β−12=0)),((α^2 −2β^2 +2c−12=0)),((α^3 +2β^3 +6βγ−12=0)) :} ⇒ { ((α=−2β+12)),((γ=−3β^2 +24β−66)),((β^3 −12β^2 +((105)/2)β−((143)/2)=0)) :} x^4 +y^4 +z^4 =δ ⇒ β^3 −12β^2 +((105)/2)β−((1227)/(16))=−(δ/(384)) ⇒ δ=1992 x^4 +y^4 +z^4 =1992

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

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

Verry Nice put y and Z=β_− ^+ (√γ)

$${Verry}\:{Nice}\:{put}\:{y}\:{and}\:{Z}=\beta_{−} ^{+} \sqrt{\gamma} \\ $$$$ \\ $$

Answered by ajfour last updated on 26/Sep/19

(x+y+z)^3 =x^3 +y^3 +z^3 + Σ3xy(12−z)+6xyz ⇒(12)^3 =12−3xyz+36Σxy ..(i) (x+y+z)^2 =Σx^2 +2Σxy ⇒ Σxy=66 now from (i) 3xyz = 3p=12+36×66−(12)^3 =12{1+192−144} =12×49 (Σx^2 )^2 =Σx^4 +2Σx^2 y^2 =Q+2{(Σxy)^2 −2pΣx} ⇒ Q=144−2{(66)^2 −8×12×49} =144−24{363−392} =144+24×29 = 24×35 =840 .

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

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