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Question Number 155377 by mathdanisur last updated on 29/Sep/21

$$\mathrm{For}\mathrm{two}\mathrm{non}-\mathrm{negative}\mathrm{real}\mathrm{numbers}:$$$${\mathrm{a}}^6+{\mathrm{b}}^6=2$$$$\mathrm{Find}\mathrm{a};\mathrm{b}\mathrm{such}\mathrm{that}3(\mathrm{a}+\mathrm{b})=1+5\sqrt{\mathrm{ab}}$$

Answered by MJS_new last updated on 30/Sep/21

$$\mathrm{the}\mathrm{only}\mathrm{real}\mathrm{solution}\mathrm{is}a=b=1$$

Commented by mathdanisur last updated on 30/Sep/21

$$\mathrm{Thankyou}\mathbf{S}\mathrm{er},\mathrm{how}\mathrm{if}\mathrm{possible}$$

Answered by MJS_new last updated on 30/Sep/21

$$3(a+b)=1+5\sqrt{ab}$$$$\mathrm{let}a=u-v\wedge b=u+v$$$$6u=1+5\sqrt{u^2-v^2}\Rightarrow v^2=-\frac{11u^2-12u+1}{25}$$$$a^6+b^6-2=0$$$$u^6+15u^4{v}^2+15u^2{v}^4+v^6-1=0$$$$\mathrm{inserting}\mathrm{\&}\mathrm{transforming}$$$$u^6-\frac{279}{679}{u}^5-\frac{2985}{2716}{u}^4+\frac{405}{2716}{u}^3+\frac{45}{21728}{u}^2-\frac{9}{10864}u+\frac{7813}{21728}=0$$$${(u-1)}^2(u^4+\frac{1079}{679}{u}^3+\frac{2931}{2716}{u}^2+\frac{1951}{2716}u+\frac{7813}{21728})=0$$$$\mathrm{this}\mathrm{has}\mathrm{no}\mathrm{other}\mathrm{real}\mathrm{solution}\mathrm{than}$$$$u=1$$$$\Rightarrow$$$$v=0$$$$\Rightarrow$$$$a=b=1$$

Commented by mathdanisur last updated on 30/Sep/21

$$\mathrm{Very}\mathrm{impressive}\mathbf{S}\mathrm{er}\mathrm{thank}\mathrm{you}$$

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