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Question Number 183592 by greougoury555 last updated on 27/Dec/22

      { (((√(x/y)) +(√(y/z)) +(√(z/x)) = 3)),(((√(y/x)) +(√(z/y)) +(√(x/z)) = 3)),(((√(xyz)) = 1)) :}

$$\:\:\:\:\:\begin{cases}{\sqrt{\frac{{x}}{{y}}}\:+\sqrt{\frac{{y}}{{z}}}\:+\sqrt{\frac{{z}}{{x}}}\:=\:\mathrm{3}}\\{\sqrt{\frac{{y}}{{x}}}\:+\sqrt{\frac{{z}}{{y}}}\:+\sqrt{\frac{{x}}{{z}}}\:=\:\mathrm{3}}\\{\sqrt{{xyz}}\:=\:\mathrm{1}}\end{cases} \\ $$$$\: \\ $$$$ \\ $$

Answered by mr W last updated on 27/Dec/22

a=(√(x/y)), b=(√(y/z)), c=(√(z/x))  abc=1  a+b+c=3  (1/a)+(1/b)+(1/c)=3  ab+bc+ca=3  a,b,c are roots of  t^3 −3t^2 +3t−1=0  (t−1)^3 =0  ⇒a=b=c=t=1  ⇒x=y=z=1 ✓

$${a}=\sqrt{\frac{{x}}{{y}}},\:{b}=\sqrt{\frac{{y}}{{z}}},\:{c}=\sqrt{\frac{{z}}{{x}}} \\ $$$${abc}=\mathrm{1} \\ $$$${a}+{b}+{c}=\mathrm{3} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{3} \\ $$$${ab}+{bc}+{ca}=\mathrm{3} \\ $$$${a},{b},{c}\:{are}\:{roots}\:{of} \\ $$$${t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow{a}={b}={c}={t}=\mathrm{1} \\ $$$$\Rightarrow{x}={y}={z}=\mathrm{1}\:\checkmark \\ $$

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