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Question Number 158303 by HongKing last updated on 02/Nov/21

x;y;z;t>0 solve for real numbers: { ((8x^4 + 64y^4 + 216z^4 + 1728t^4 = 1)),((x + y + z + t = 1)) :}

$$\mathrm{x};\mathrm{y};\mathrm{z};\mathrm{t}>\mathrm{0} \\ $$ $$\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$ $$\begin{cases}{\mathrm{8x}^{\mathrm{4}} \:+\:\mathrm{64y}^{\mathrm{4}} \:+\:\mathrm{216z}^{\mathrm{4}} \:+\:\mathrm{1728t}^{\mathrm{4}} \:=\:\mathrm{1}}\\{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:+\:\mathrm{t}\:=\:\mathrm{1}}\end{cases} \\ $$ $$ \\ $$

Answered by mr W last updated on 02/Nov/21

(((2x)^4 )/2)+(((4y)^4 )/4)+(((6z)^4 )/6)+(((12t)^4 )/(12))=1 ((2x)/2)+((4y)/4)+((6z)/6)+((12t)/(12))=1 (a^4 /2)+(b^4 /4)+(c^4 /6)+(d^4 /(12))=1 (a/2)+(b/4)+(c/6)+(d/(12))=1 ⇒a=b=c=d=1 ⇒x=(1/2), y=(1/4), z=(1/6), t=(1/(12))

$$\frac{\left(\mathrm{2}{x}\right)^{\mathrm{4}} }{\mathrm{2}}+\frac{\left(\mathrm{4}{y}\right)^{\mathrm{4}} }{\mathrm{4}}+\frac{\left(\mathrm{6}{z}\right)^{\mathrm{4}} }{\mathrm{6}}+\frac{\left(\mathrm{12}{t}\right)^{\mathrm{4}} }{\mathrm{12}}=\mathrm{1} \\ $$ $$\frac{\mathrm{2}{x}}{\mathrm{2}}+\frac{\mathrm{4}{y}}{\mathrm{4}}+\frac{\mathrm{6}{z}}{\mathrm{6}}+\frac{\mathrm{12}{t}}{\mathrm{12}}=\mathrm{1} \\ $$ $$\frac{{a}^{\mathrm{4}} }{\mathrm{2}}+\frac{{b}^{\mathrm{4}} }{\mathrm{4}}+\frac{{c}^{\mathrm{4}} }{\mathrm{6}}+\frac{{d}^{\mathrm{4}} }{\mathrm{12}}=\mathrm{1} \\ $$ $$\frac{{a}}{\mathrm{2}}+\frac{{b}}{\mathrm{4}}+\frac{{c}}{\mathrm{6}}+\frac{{d}}{\mathrm{12}}=\mathrm{1} \\ $$ $$\Rightarrow{a}={b}={c}={d}=\mathrm{1} \\ $$ $$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}},\:{y}=\frac{\mathrm{1}}{\mathrm{4}},\:{z}=\frac{\mathrm{1}}{\mathrm{6}},\:{t}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Commented byHongKing last updated on 02/Nov/21

cool my dear Sir, thank you so much

$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{ir},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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