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Question Number 132494 by mnjuly1970 last updated on 14/Feb/21

Commented by MJS_new last updated on 14/Feb/21

sharing these transformations we have to check all solutions!!! a^(1/2) +b^(1/2) =c^(1/2) ∣^2 and transforming 2a^(1/2) b^(1/2) =−a−b+c ∣^2 4ab=(a+b−c)^2 a^(1/3) +b^(1/3) =c^(1/3) ∣^3 and transforming 3a^(1/3) b^(1/3) (a^(1/3) +b^(1/3) )=−a−b+c but a^(1/3) +b^(1/3) =c^(1/3) 3a^(1/3) b^(1/3) c^(1/3) =−a−b+c ∣^3 27abc=−(a+b−c)^3 a^(1/4) +b^(1/4) =c^(1/4) starting with result of the square root equation sbove 4ab=(a+b−c)^2 ⇔ a^2 +b^2 +c^2 −2(ab+ac+bc)=0 ⇒ a+b+c−2(a^(1/2) b^(1/2) +a^(1/2) c^(1/2) +b^(1/2) c^(1/2) )=0 transforming a+b+c−2a^(1/2) b^(1/2) =2c^(1/2) (a^(1/2) +b^(1/2) ) ∣^2 and transforming a^2 +b^2 +c^2 −6ab−2ac−2bc=4a^(1/2) b^(1/2) (a+b+3c) ∣^2 and transforming 8ab(a^2 +b^2 +17c^2 +14ac+14bc)=(a+b−c)^4 I like these: (((a+b−c)^2 )/(4ab))=1 (((a+b−c)^3 )/(27abc))=−1 (((a+b−c)^4 )/(8ab))=a^2 +b^2 +17c^2 +14(a+b)c

$$\mathrm{sharing}\:\mathrm{these}\:\mathrm{transformations} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{check}\:\mathrm{all}\:\mathrm{solutions}!!! \\ $$$$ \\ $$$${a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} ={c}^{\mathrm{1}/\mathrm{2}} \\ $$$$\mid^{\mathrm{2}} \:\mathrm{and}\:\mathrm{transforming} \\ $$$$\mathrm{2}{a}^{\mathrm{1}/\mathrm{2}} {b}^{\mathrm{1}/\mathrm{2}} =−{a}−{b}+{c} \\ $$$$\mid^{\mathrm{2}} \\ $$$$\mathrm{4}{ab}=\left({a}+{b}−{c}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${a}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{1}/\mathrm{3}} ={c}^{\mathrm{1}/\mathrm{3}} \\ $$$$\mid^{\mathrm{3}} \:\mathrm{and}\:\mathrm{transforming} \\ $$$$\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} \left({a}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{1}/\mathrm{3}} \right)=−{a}−{b}+{c} \\ $$$$\mathrm{but}\:{a}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{1}/\mathrm{3}} ={c}^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} {c}^{\mathrm{1}/\mathrm{3}} =−{a}−{b}+{c} \\ $$$$\mid^{\mathrm{3}} \\ $$$$\mathrm{27}{abc}=−\left({a}+{b}−{c}\right)^{\mathrm{3}} \\ $$$$ \\ $$$${a}^{\mathrm{1}/\mathrm{4}} +{b}^{\mathrm{1}/\mathrm{4}} ={c}^{\mathrm{1}/\mathrm{4}} \\ $$$$\mathrm{starting}\:\mathrm{with}\:\mathrm{result}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{root} \\ $$$$\mathrm{equation}\:\mathrm{sbove} \\ $$$$\mathrm{4ab}=\left(\mathrm{a}+\mathrm{b}−\mathrm{c}\right)^{\mathrm{2}} \:\Leftrightarrow\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${a}+{b}+{c}−\mathrm{2}\left({a}^{\mathrm{1}/\mathrm{2}} {b}^{\mathrm{1}/\mathrm{2}} +{a}^{\mathrm{1}/\mathrm{2}} {c}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} {c}^{\mathrm{1}/\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{transforming} \\ $$$${a}+{b}+{c}−\mathrm{2}{a}^{\mathrm{1}/\mathrm{2}} {b}^{\mathrm{1}/\mathrm{2}} =\mathrm{2}{c}^{\mathrm{1}/\mathrm{2}} \left({a}^{\mathrm{1}/\mathrm{2}} +{b}^{\mathrm{1}/\mathrm{2}} \right) \\ $$$$\mid^{\mathrm{2}} \:\mathrm{and}\:\mathrm{transforming} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{6}{ab}−\mathrm{2}{ac}−\mathrm{2}{bc}=\mathrm{4}{a}^{\mathrm{1}/\mathrm{2}} {b}^{\mathrm{1}/\mathrm{2}} \left({a}+{b}+\mathrm{3}{c}\right) \\ $$$$\mid^{\mathrm{2}} \:\mathrm{and}\:\mathrm{transforming} \\ $$$$\mathrm{8}{ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{17}{c}^{\mathrm{2}} +\mathrm{14}{ac}+\mathrm{14}{bc}\right)=\left({a}+{b}−{c}\right)^{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{like}\:\mathrm{these}: \\ $$$$\frac{\left({a}+{b}−{c}\right)^{\mathrm{2}} }{\mathrm{4}{ab}}=\mathrm{1} \\ $$$$\frac{\left({a}+{b}−{c}\right)^{\mathrm{3}} }{\mathrm{27}{abc}}=−\mathrm{1} \\ $$$$\frac{\left({a}+{b}−{c}\right)^{\mathrm{4}} }{\mathrm{8}{ab}}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{17}{c}^{\mathrm{2}} +\mathrm{14}\left({a}+{b}\right){c} \\ $$

Commented by MJS_new last updated on 14/Feb/21

I have these up to a^(1/9) +b^(1/9) =c^(1/9) but I forgot the paths...

$$\mathrm{I}\:\mathrm{have}\:\mathrm{these}\:\mathrm{up}\:\mathrm{to}\:{a}^{\mathrm{1}/\mathrm{9}} +{b}^{\mathrm{1}/\mathrm{9}} ={c}^{\mathrm{1}/\mathrm{9}} \:\mathrm{but}\:\mathrm{I}\:\mathrm{forgot} \\ $$$$\mathrm{the}\:\mathrm{paths}... \\ $$

Commented by mnjuly1970 last updated on 14/Feb/21

grateful mr mjs...

$${grateful}\:{mr}\:{mjs}... \\ $$

Answered by MJS_new last updated on 14/Feb/21

using the 3^(rd) above equation with a=x−1∧b=x+1∧c=x+3 we get x^4 +4x^3 +((54)/(25))x^2 −((92)/(25))x−((1321)/(375))=0 let x=z−1 z^4 −((96)/(25))z^2 −((256)/(375))=0 and this is easy to solve of the 4 solutions only x=((−15+(√(3(9+4(√6)))))/(15)) solves the given equation

$$\mathrm{using}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{above}\:\mathrm{equation}\:\mathrm{with} \\ $$$${a}={x}−\mathrm{1}\wedge{b}={x}+\mathrm{1}\wedge{c}={x}+\mathrm{3} \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\frac{\mathrm{54}}{\mathrm{25}}{x}^{\mathrm{2}} −\frac{\mathrm{92}}{\mathrm{25}}{x}−\frac{\mathrm{1321}}{\mathrm{375}}=\mathrm{0} \\ $$$$\mathrm{let}\:{x}={z}−\mathrm{1} \\ $$$${z}^{\mathrm{4}} −\frac{\mathrm{96}}{\mathrm{25}}{z}^{\mathrm{2}} −\frac{\mathrm{256}}{\mathrm{375}}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{this}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{4}\:\mathrm{solutions}\:\mathrm{only} \\ $$$${x}=\frac{−\mathrm{15}+\sqrt{\mathrm{3}\left(\mathrm{9}+\mathrm{4}\sqrt{\mathrm{6}}\right)}}{\mathrm{15}} \\ $$$$\mathrm{solves}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$

Answered by Tawa11 last updated on 06/Nov/21

Great

$$\mathrm{Great} \\ $$

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