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Question Number 201802 by Frix last updated on 12/Dec/23

Solution of equations like this: ((f(x)))^(1/n) +((g(x)))^(1/n) =((h(x)))^(1/n) If the solution is not obvious we must get rid of the radicals. In the following cases this is easy: (√a)+(√b)=(√c) ⇒ a+2(√(ab))+b=c ⇒ 4ab=(c−a−b)^2 (a)^(1/3) +(b)^(1/3) =(c)^(1/3) ⇒ a+3((ab))^(1/3) ((a)^(1/3) +(b)^(1/3) )+b=c ⇒ a+3((abc))^(1/3) +b=c ⇒ 27abc=(c−a−b)^3 But how to solve for n≥4? I found this formula to get an equation without radicals: a^(1/n) +b^(1/n) =c^(1/n) ⇒ Π_(k=0) ^(n−1) (c−(a^(1/n) +b^(1/n) e^(i((2πk)/n)) )^n ) =0 For n=2, 3 we get above equations. For n=4: c^4 −4(a+b)c^3 +2(3a^2 −62ab+3b^2 )c^2 −4(a+b)(a^2 +30ab+b^2 )c+(a−b)^4 =0 ⇔ 8ab(17c^2 +14(a+b)c+a^2 +b^2 )=(c−a−b)^4 For n=5: 625abc(c^2 +3(a+b)c+a^2 −3ab+b^2 )=(c−a−b)^5 I hope this helps...

$$\mathrm{Solution}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{like}\:\mathrm{this}: \\ $$$$\sqrt[{{n}}]{{f}\left({x}\right)}+\sqrt[{{n}}]{{g}\left({x}\right)}=\sqrt[{{n}}]{{h}\left({x}\right)} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{not}\:\mathrm{obvious}\:\mathrm{we}\:\mathrm{must}\:\mathrm{get} \\ $$$$\mathrm{rid}\:\mathrm{of}\:\mathrm{the}\:\mathrm{radicals}.\:\mathrm{In}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{easy}: \\ $$$$\sqrt{{a}}+\sqrt{{b}}=\sqrt{{c}} \\ $$$$\:\:\:\:\:\Rightarrow\:{a}+\mathrm{2}\sqrt{{ab}}+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{4}{ab}=\left({c}−{a}−{b}\right)^{\mathrm{2}} \\ $$$$\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}=\sqrt[{\mathrm{3}}]{{c}} \\ $$$$\:\:\:\:\:\Rightarrow\:{a}+\mathrm{3}\sqrt[{\mathrm{3}}]{{ab}}\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}\right)+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{a}+\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{27}{abc}=\left({c}−{a}−{b}\right)^{\mathrm{3}} \\ $$$$\mathrm{But}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for}\:{n}\geqslant\mathrm{4}? \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{this}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{get}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{without}\:\mathrm{radicals}: \\ $$$${a}^{\frac{\mathrm{1}}{{n}}} +{b}^{\frac{\mathrm{1}}{{n}}} ={c}^{\frac{\mathrm{1}}{{n}}} \\ $$$$\Rightarrow \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\:\left({c}−\left({a}^{\frac{\mathrm{1}}{{n}}} +{b}^{\frac{\mathrm{1}}{{n}}} \mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi{k}}{{n}}} \right)^{{n}} \right)\:=\mathrm{0} \\ $$$$\mathrm{For}\:{n}=\mathrm{2},\:\mathrm{3}\:\mathrm{we}\:\mathrm{get}\:\mathrm{above}\:\mathrm{equations}. \\ $$$$\mathrm{For}\:{n}=\mathrm{4}: \\ $$$${c}^{\mathrm{4}} −\mathrm{4}\left({a}+{b}\right){c}^{\mathrm{3}} +\mathrm{2}\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{62}{ab}+\mathrm{3}{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} −\mathrm{4}\left({a}+{b}\right)\left({a}^{\mathrm{2}} +\mathrm{30}{ab}+{b}^{\mathrm{2}} \right){c}+\left({a}−{b}\right)^{\mathrm{4}} =\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{8}{ab}\left(\mathrm{17}{c}^{\mathrm{2}} +\mathrm{14}\left({a}+{b}\right){c}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\left({c}−{a}−{b}\right)^{\mathrm{4}} \\ $$$$\mathrm{For}\:{n}=\mathrm{5}: \\ $$$$\mathrm{625}{abc}\left({c}^{\mathrm{2}} +\mathrm{3}\left({a}+{b}\right){c}+{a}^{\mathrm{2}} −\mathrm{3}{ab}+{b}^{\mathrm{2}} \right)=\left({c}−{a}−{b}\right)^{\mathrm{5}} \\ $$$$\mathrm{I}\:\mathrm{hope}\:\mathrm{this}\:\mathrm{helps}... \\ $$

Commented by Calculusboy last updated on 12/Dec/23

yes

$$\boldsymbol{{yes}} \\ $$

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