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x-abc-1-4-x-abc-1-8-a-x-abc-1-4-x-abc-1-8-b-x-abc-1-4-x-abc-1-4-c-find-x-abc-x-abc-

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Question Number 161065 by cortano last updated on 11/Dec/21
{ ((((x+abc))^(1/4) +((x−abc))^(1/8) = a)),((((x+abc))^(1/4) −((x−abc))^(1/8) = b)),((((x+abc))^(1/4) −((x−abc))^(1/4) = c)) :} find (√(x+abc)) +(√(x−abc))
$$\:\begin{cases}{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:+\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{a}}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{b}}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{4}}]{{x}−{abc}}\:=\:{c}}\end{cases} \\ $$$$\:{find}\:\sqrt{{x}+{abc}}\:+\sqrt{{x}−{abc}} \\ $$
Answered by Rasheed.Sindhi last updated on 12/Dec/21
{ ((((x+abc))^(1/4) +((x−abc))^(1/8) = a....(i))),((((x+abc))^(1/4) −((x−abc))^(1/8) = b.....(ii))),((((x+abc))^(1/4) −((x−abc))^(1/4) = c.....(iii))) :} 2((x+abc))^(1/4) =a+b ((x+abc))^(1/4) =((a+b)/2) (((x+abc))^(1/4) )^2 =(((a+b)/2))^2 (√(x+abc)) =(((a+b)/2))^2 ...........A 2((x−abc))^(1/8) =a−b ((x−abc))^(1/8) =((a−b)/2) (((x−abc))^(1/8) )^4 =(((a−b)/2))^4 (√(x−abc)) =(((a−b)/2))^4 .........B A+B: (√(x+abc)) +(√(x−abc)) =(((a+b)/2))^2 +(((a−b)/2))^4 Without using (iii) !!!
$$\:\begin{cases}{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:+\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{a}….\left({i}\right)}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{b}…..\left({ii}\right)}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{4}}]{{x}−{abc}}\:=\:{c}…..\left({iii}\right)}\end{cases} \\ $$$$\mathrm{2}\sqrt[{\mathrm{4}}]{{x}+{abc}}\:={a}+{b} \\ $$$$\sqrt[{\mathrm{4}}]{{x}+{abc}}\:=\frac{{a}+{b}}{\mathrm{2}} \\ $$$$\left(\sqrt[{\mathrm{4}}]{{x}+{abc}}\:\right)^{\mathrm{2}} =\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\sqrt{{x}+{abc}}\:=\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} ………..{A} \\ $$$$\mathrm{2}\sqrt[{\mathrm{8}}]{{x}−{abc}}\:={a}−{b} \\ $$$$\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\frac{{a}−{b}}{\mathrm{2}} \\ $$$$\left(\sqrt[{\mathrm{8}}]{{x}−{abc}}\:\right)^{\mathrm{4}} =\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{4}} \\ $$$$\sqrt{{x}−{abc}}\:=\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{4}} ………{B} \\ $$$${A}+{B}:\:\sqrt{{x}+{abc}}\:+\sqrt{{x}−{abc}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}−{b}}{\mathrm{2}}\right)^{\mathrm{4}} \\ $$$${Without}\:{using}\:\left({iii}\right)\:!!! \\ $$

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