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Question Number 154804 by mathdanisur last updated on 21/Sep/21

Solve for real numbers: ((5(√5) + x))^(1/5) - ((5(√5) - x))^(1/5) = (2)^(1/5)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt[{\mathrm{5}}]{\mathrm{5}\sqrt{\mathrm{5}}\:+\:\mathrm{x}}\:-\:\sqrt[{\mathrm{5}}]{\mathrm{5}\sqrt{\mathrm{5}}\:-\:\mathrm{x}}\:=\:\sqrt[{\mathrm{5}}]{\mathrm{2}} \\ $$

Commented by 7770 last updated on 21/Sep/21

x=11

$$\mathrm{x}=\mathrm{11} \\ $$

Answered by TheHoneyCat last updated on 21/Sep/21

let f: { (R,→,R),(x, ,((5(√5)+x)^(1/5) )) :} Re−writting the equation we get: (2)^(1/5) =(5(√5)+x)^(1/5) −(5(√5)−x)^(1/5) =f(x)−f(−x) very obviously f∈D^1 (R) as a composition of C^∞ (R) ones actually, to be rigorous R sould be replaced by R−{5(√5)} cause ( )^(1/n) is never derivable in 0... but you get the point (df/dx)=((d(5(√5)+x))/dx)×((d(x^(1/5) ))/dx)○(x 5(√5)+x) =x (1/5)(5(√5)+x)^(−4/5) =x (1/5)((5(√5)+x)^2 )^(−2/5) ∀x∈R−{−5(√5)} (5(√5)+x)^2 >0 so ∀x∈R−{−5(√5)} ((5(√5)+x)^2 )^(−2/5) >0 hence (df/(dx ))>0 (over R−{5(√5)} of course this gives (df/dx)○(x −x)<0 (over R−{−5(√5)}) since f and (therefor x f(x)−f(−x) ) is strictly increasing it as 1 solution or less considering x=11 ((5^(3/2) +11)^(1/5) −(5^(3/2) −11)^(1/5) )^5 by devellopping with newton′s formula the“^5 ” you get 2. its a pain, so I won′t show it here. Sorry Hence x=11_■

$$\mathrm{let}\:{f}:\begin{cases}{\mathbb{R}}&{\rightarrow}&{\mathbb{R}}\\{{x}}&{ }&{\left(\mathrm{5}\sqrt{\mathrm{5}}+{x}\right)^{\mathrm{1}/\mathrm{5}} }\end{cases} \\ $$$$ \\ $$$$\mathrm{Re}−\mathrm{writting}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{get}: \\ $$$$\sqrt[{\mathrm{5}}]{\mathrm{2}}=\left(\mathrm{5}\sqrt{\mathrm{5}}+{x}\right)^{\mathrm{1}/\mathrm{5}} −\left(\mathrm{5}\sqrt{\mathrm{5}}−{x}\right)^{\mathrm{1}/\mathrm{5}} \\ $$$$={f}\left({x}\right)−{f}\left(−{x}\right) \\ $$$$ \\ $$$$\mathrm{very}\:\mathrm{obviously}\:{f}\in\mathscr{D}^{\mathrm{1}} \left(\mathbb{R}\right)\:\mathrm{as}\:\mathrm{a}\:\mathrm{composition}\:\mathrm{of}\:\mathscr{C}^{\infty} \left(\mathbb{R}\right)\:\mathrm{ones} \\ $$$${actually},\:{to}\:{be}\:{rigorous}\:\mathbb{R}\:{sould}\:{be}\:{replaced}\:{by}\:\mathbb{R}−\left\{\mathrm{5}\sqrt{\mathrm{5}}\right\}\:{cause}\:\sqrt[{{n}}]{\:}\:{is}\:{never}\:{derivable}\:{in}\:\mathrm{0}...\:{but}\:{you}\:{get}\:{the}\:{point} \\ $$$$\frac{\mathrm{d}{f}}{\mathrm{d}{x}}=\frac{\mathrm{d}\left(\mathrm{5}\sqrt{\mathrm{5}}+{x}\right)}{\mathrm{d}{x}}×\frac{\mathrm{d}\left({x}^{\mathrm{1}/\mathrm{5}} \right)}{\mathrm{d}{x}}\circ\left({x} \mathrm{5}\sqrt{\mathrm{5}}+{x}\right) \\ $$$$={x} \frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{5}\sqrt{\mathrm{5}}+{x}\right)^{−\mathrm{4}/\mathrm{5}} \\ $$$$={x} \frac{\mathrm{1}}{\mathrm{5}}\left(\left(\mathrm{5}\sqrt{\mathrm{5}}+{x}\right)^{\mathrm{2}} \right)^{−\mathrm{2}/\mathrm{5}} \\ $$$$ \\ $$$$\forall{x}\in\mathbb{R}−\left\{−\mathrm{5}\sqrt{\mathrm{5}}\right\}\:\left(\mathrm{5}\sqrt{\mathrm{5}}+{x}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{so}\:\forall{x}\in\mathbb{R}−\left\{−\mathrm{5}\sqrt{\mathrm{5}}\right\}\:\left(\left(\mathrm{5}\sqrt{\mathrm{5}}+{x}\right)^{\mathrm{2}} \right)^{−\mathrm{2}/\mathrm{5}} >\mathrm{0} \\ $$$$\mathrm{hence}\:\frac{\mathrm{d}{f}}{\mathrm{d}{x}\:}>\mathrm{0}\:\left({over}\:\mathbb{R}−\left\{\mathrm{5}\sqrt{\mathrm{5}}\right\}\right. \\ $$$$\mathrm{of}\:\mathrm{course}\:\mathrm{this}\:\mathrm{gives}\:\frac{\mathrm{d}{f}}{\mathrm{d}{x}}\circ\left({x} −{x}\right)<\mathrm{0}\:\left({over}\:\mathbb{R}−\left\{−\mathrm{5}\sqrt{\mathrm{5}}\right\}\right) \\ $$$$ \\ $$$$\mathrm{since}\:{f}\:\mathrm{and}\:\left(\mathrm{therefor}\:{x} {f}\left({x}\right)−{f}\left(−{x}\right)\:\right)\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{increasing} \\ $$$$\mathrm{it}\:\mathrm{as}\:\mathrm{1}\:\mathrm{solution}\:\mathrm{or}\:\mathrm{less} \\ $$$$ \\ $$$$\mathrm{considering}\:{x}=\mathrm{11} \\ $$$$\left(\left(\mathrm{5}^{\mathrm{3}/\mathrm{2}} +\mathrm{11}\right)^{\mathrm{1}/\mathrm{5}} −\left(\mathrm{5}^{\mathrm{3}/\mathrm{2}} −\mathrm{11}\right)^{\mathrm{1}/\mathrm{5}} \right)^{\mathrm{5}} \\ $$$$\mathrm{by}\:\mathrm{devellopping}\:\mathrm{with}\:\mathrm{newton}'\mathrm{s}\:\mathrm{formula}\:\mathrm{the}``\:^{\mathrm{5}} ''\:\mathrm{you}\:\mathrm{get}\:\mathrm{2}. \\ $$$${its}\:{a}\:{pain},\:{so}\:{I}\:{won}'{t}\:{show}\:{it}\:{here}.\:{Sorry} \\ $$$$ \\ $$$$\mathrm{Hence}\:{x}=\mathrm{11}_{\blacksquare} \\ $$

Commented by TheHoneyCat last updated on 21/Sep/21

since the calculation was “a pain” you might wonder how I found that x=11 would work in the first place did I stubble upon it by pure luck? well, not at all, actualy this whole proof is the opposite of what I did by myself I actualy plotted the curve on a graph and found (to a very good precision) that the only solution was indeed 11 I then just had to proove the unicity and verify that 11 was not just an approximation see the folowing pictures I plotted.

$${since}\:{the}\:{calculation}\:{was}\:``{a}\:{pain}''\:{you}\:{might}\:{wonder}\:{how}\:{I}\:{found}\:{that}\:{x}=\mathrm{11}\:{would}\:{work}\:{in}\:{the}\:{first}\:{place} \\ $$$${did}\:{I}\:{stubble}\:{upon}\:{it}\:{by}\:{pure}\:{luck}? \\ $$$${well},\:{not}\:{at}\:{all},\:{actualy}\:{this}\:{whole}\:{proof}\:{is}\:{the}\:{opposite}\:{of}\:{what}\:{I}\:{did}\:{by}\:{myself} \\ $$$${I}\:{actualy}\:{plotted}\:{the}\:{curve}\:{on}\:{a}\:{graph}\:{and}\:{found}\:\left({to}\:{a}\:{very}\:{good}\:{precision}\right)\:{that}\:{the}\:{only}\:{solution}\:{was}\:{indeed}\:\mathrm{11} \\ $$$${I}\:{then}\:{just}\:{had}\:{to}\:{proove}\:{the}\:{unicity}\:{and}\:{verify}\:{that}\:\mathrm{11}\:{was}\:{not}\:{just}\:{an}\:{approximation} \\ $$$${see}\:{the}\:{folowing}\:{pictures}\:{I}\:{plotted}. \\ $$

Commented by TheHoneyCat last updated on 21/Sep/21

Here is the equation (twisted to get an “=0” statement)

$$\mathrm{Here}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equation}\:\left({twisted}\:{to}\:{get}\:{an}\:``=\mathrm{0}''\:{statement}\right) \\ $$

Commented by TheHoneyCat last updated on 21/Sep/21

Commented by TheHoneyCat last updated on 21/Sep/21

and here is the derivative (I wonted to make sure it was indeed >0 if it hadn′t been, I would have not tried trought this method)

$$\mathrm{and}\:\mathrm{here}\:\mathrm{is}\:\mathrm{the}\:\mathrm{derivative} \\ $$$$\left({I}\:{wonted}\:{to}\:{make}\:{sure}\:{it}\:{was}\:{indeed}\:>\mathrm{0}\right. \\ $$$$\left.{if}\:{it}\:{hadn}'{t}\:{been},\:{I}\:{would}\:{have}\:{not}\:{tried}\:{trought}\:{this}\:{method}\right) \\ $$

Commented by TheHoneyCat last updated on 21/Sep/21

Commented by TheHoneyCat last updated on 21/Sep/21

Now, here′s a real hard question, both geogebra and maple, when asked to solve this, give an exact value (11) But neither one of them does tel me how they found that it was an exact value Does this means they are just too lazy to tel me, but there DOES exist a way to find it without first knowing it (ie without a computer)? Or are they just pretending that the value is exact because it is coded in such a way that 11.000000000000000001 is showed as 11 (ie the value shown is not formalay calculated, and its just a float) This might be a very “metha” question for it is at the fronteer of math, computer science and raises the (almost philosophical) question : what does it mean for an equation to be solvable ? Please if you have the answere, do give it, this seems super interesting :−)

$$\mathrm{Now},\:\mathrm{here}'\mathrm{s}\:\mathrm{a}\:\mathrm{real}\:\mathrm{hard}\:\mathrm{question}, \\ $$$$\mathrm{both}\:\mathrm{geogebra}\:\mathrm{and}\:\mathrm{maple},\:\mathrm{when}\:\mathrm{asked}\: \\ $$$$\mathrm{to}\:\mathrm{solve}\:\mathrm{this},\:\mathrm{give}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{value}\:\left(\mathrm{11}\right) \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{neither}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them}\:\mathrm{does}\:\mathrm{tel}\:\mathrm{me}\:\mathrm{how} \\ $$$$\mathrm{they}\:\mathrm{found}\:\mathrm{that}\:\mathrm{it}\:\mathrm{was}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{value} \\ $$$$ \\ $$$$\mathrm{Does}\:\mathrm{this}\:\mathrm{means}\:\mathrm{they}\:\mathrm{are}\:\mathrm{just}\:\mathrm{too}\:\mathrm{lazy}\:\mathrm{to} \\ $$$$\mathrm{tel}\:\mathrm{me},\:\mathrm{but}\:\mathrm{there}\:\mathrm{DOES}\:\mathrm{exist}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{find}\: \\ $$$$\mathrm{it}\:\mathrm{without}\:\mathrm{first}\:\mathrm{knowing}\:\mathrm{it}\:\left({ie}\:{without}\:{a}\:\right. \\ $$$$\left.{computer}\right)?\:\mathrm{Or}\:\mathrm{are}\:\mathrm{they}\:\mathrm{just}\:\mathrm{pretending} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{value}\:\mathrm{is}\:\mathrm{exact}\:\mathrm{because}\:\mathrm{it}\:\mathrm{is}\:\mathrm{coded} \\ $$$$\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{11}.\mathrm{000000000000000001} \\ $$$$\mathrm{is}\:\mathrm{showed}\:\mathrm{as}\:\mathrm{11}\:\left({ie}\:{the}\:{value}\:{shown}\:{is}\:{not}\right. \\ $$$$\left.{formalay}\:{calculated},\:{and}\:{its}\:{just}\:{a}\:{float}\right) \\ $$$$ \\ $$$$\mathrm{This}\:\mathrm{might}\:\mathrm{be}\:\mathrm{a}\:\mathrm{very}\:``{metha}''\:\mathrm{question}\: \\ $$$$\mathrm{for}\:\mathrm{it}\:\mathrm{is}\:\mathrm{at}\:\mathrm{the}\:\mathrm{fronteer}\:\mathrm{of}\:\mathrm{math},\:\mathrm{computer}\:\mathrm{science} \\ $$$$\mathrm{and}\:\mathrm{raises}\:\mathrm{the}\:\left({almost}\:{philosophical}\right)\: \\ $$$$\mathrm{question}\::\:\mathrm{what}\:\mathrm{does}\:\mathrm{it}\:\mathrm{mean}\:\mathrm{for}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{solvable}\:? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{if}\:\mathrm{you}\:\mathrm{have}\:\mathrm{the}\:\mathrm{answere},\:\mathrm{do}\:\mathrm{give}\:\mathrm{it}, \\ $$$$\mathrm{this}\:\mathrm{seems}\:\mathrm{super}\:\mathrm{interesting} \\ $$$$\left.:−\right) \\ $$

Commented by mathdanisur last updated on 21/Sep/21

let a=5(√5)+x ; b=5(√5)-x then a-b=(2)^(1/5) ; a^5 -b^5 =2x and a^5 +b^5 =10(√5) then (a/( (2)^(1/5) ))-(b/( (2)^(1/5) ))=1 ; ((a/( (2)^(1/5) )))^5 -((b/( (2)^(1/5) )))^5 =x and ((a/( (2)^(1/5) )))^5 +((b/( (2)^(1/5) )))^5 =5(√5) let (a/( (2)^(1/5) ))=p and (b/( (2)^(1/5) ))=q we have p-q=1 (1) ; p^5 -q^5 =x (2) ; p^5 +q^5 =5(√5) (3) also we knov p^5 +q^5 =(p+q)[(p-q)^2 ((p-q)^2 +4pq)+2p^2 q^2 -pq((p-q)^2 +pq)] now 5(√5)=(p+q)(p^2 q^2 +3pq+1) squaring 125=(p+q)^2 (p^2 q^2 +3pq+1)^2 5^3 =(1+4pq)(p^2 q^2 +3pq+1)^2 ⇔ pq=1 doing (3)^2 -(2)^2 ⇒ 4(pq)^5 =125-x^2 ⇔ x^2 =121 ⇔ x=±11 since x=-11 does not satisfy we get x=11

$$\mathrm{let}\:\:\mathrm{a}=\mathrm{5}\sqrt{\mathrm{5}}+\mathrm{x}\:\:;\:\:\mathrm{b}=\mathrm{5}\sqrt{\mathrm{5}}-\mathrm{x} \\ $$$$\mathrm{then}\:\:\mathrm{a}-\mathrm{b}=\sqrt[{\mathrm{5}}]{\mathrm{2}}\:\:;\:\:\mathrm{a}^{\mathrm{5}} -\mathrm{b}^{\mathrm{5}} =\mathrm{2x}\:\:\mathrm{and}\:\:\mathrm{a}^{\mathrm{5}} +\mathrm{b}^{\mathrm{5}} =\mathrm{10}\sqrt{\mathrm{5}} \\ $$$$\mathrm{then}\:\:\frac{\mathrm{a}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}-\frac{\mathrm{b}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}=\mathrm{1}\:\:;\:\:\left(\frac{\mathrm{a}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}\right)^{\mathrm{5}} -\left(\frac{\mathrm{b}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}\right)^{\mathrm{5}} =\mathrm{x}\:\:\mathrm{and}\:\:\left(\frac{\mathrm{a}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}\right)^{\mathrm{5}} +\left(\frac{\mathrm{b}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}\right)^{\mathrm{5}} =\mathrm{5}\sqrt{\mathrm{5}} \\ $$$$\mathrm{let}\:\:\frac{\mathrm{a}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}=\mathrm{p}\:\:\mathrm{and}\:\:\frac{\mathrm{b}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}=\mathrm{q} \\ $$$$\mathrm{we}\:\mathrm{have}\:\:\mathrm{p}-\mathrm{q}=\mathrm{1}\:\left(\mathrm{1}\right)\:;\:\mathrm{p}^{\mathrm{5}} -\mathrm{q}^{\mathrm{5}} =\mathrm{x}\:\left(\mathrm{2}\right)\:;\:\mathrm{p}^{\mathrm{5}} +\mathrm{q}^{\mathrm{5}} =\mathrm{5}\sqrt{\mathrm{5}}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{also}\:\mathrm{we}\:\mathrm{knov}\:\:\mathrm{p}^{\mathrm{5}} +\mathrm{q}^{\mathrm{5}} =\left(\mathrm{p}+\mathrm{q}\right)\left[\left(\mathrm{p}-\mathrm{q}\right)^{\mathrm{2}} \left(\left(\mathrm{p}-\mathrm{q}\right)^{\mathrm{2}} +\mathrm{4pq}\right)+\mathrm{2p}^{\mathrm{2}} \mathrm{q}^{\mathrm{2}} -\mathrm{pq}\left(\left(\mathrm{p}-\mathrm{q}\right)^{\mathrm{2}} +\mathrm{pq}\right)\right] \\ $$$$\mathrm{now}\:\:\mathrm{5}\sqrt{\mathrm{5}}=\left(\mathrm{p}+\mathrm{q}\right)\left(\mathrm{p}^{\mathrm{2}} \mathrm{q}^{\mathrm{2}} +\mathrm{3pq}+\mathrm{1}\right) \\ $$$$\mathrm{squaring}\:\:\mathrm{125}=\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}} \left(\mathrm{p}^{\mathrm{2}} \mathrm{q}^{\mathrm{2}} +\mathrm{3pq}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}^{\mathrm{3}} =\left(\mathrm{1}+\mathrm{4pq}\right)\left(\mathrm{p}^{\mathrm{2}} \mathrm{q}^{\mathrm{2}} +\mathrm{3pq}+\mathrm{1}\right)^{\mathrm{2}} \:\Leftrightarrow\:\mathrm{pq}=\mathrm{1} \\ $$$$\mathrm{doing}\:\:\left(\mathrm{3}\right)^{\mathrm{2}} -\left(\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{4}\left(\mathrm{pq}\right)^{\mathrm{5}} =\mathrm{125}-\mathrm{x}^{\mathrm{2}} \:\Leftrightarrow\:\mathrm{x}^{\mathrm{2}} =\mathrm{121}\:\Leftrightarrow\:\mathrm{x}=\pm\mathrm{11} \\ $$$$\mathrm{since}\:\:\mathrm{x}=-\mathrm{11}\:\:\mathrm{does}\:\mathrm{not}\:\mathrm{satisfy} \\ $$$$\mathrm{we}\:\mathrm{get}\:\:\mathrm{x}=\mathrm{11} \\ $$

Commented by mathdanisur last updated on 21/Sep/21

Dear Ser, thanks for the amazing and interesting questions

$$\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{amazing}\:\mathrm{and} \\ $$$$\mathrm{interesting}\:\mathrm{questions} \\ $$

Commented by MJS_new last updated on 22/Sep/21

it′s not a proper proof in line 9 you get pq=1 but p=(a/( (2)^(1/5) ))=((11+5(√5))/( (2)^(1/5) ))∧q=(b/( (2)^(1/5) ))=((−11+5(√5))/( (2)^(1/5) )) ⇒ pq=2^(8/5) ≠1

$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{a}\:\mathrm{proper}\:\mathrm{proof} \\ $$$$\mathrm{in}\:\mathrm{line}\:\mathrm{9}\:\mathrm{you}\:\mathrm{get}\:{pq}=\mathrm{1} \\ $$$$\mathrm{but}\:{p}=\frac{{a}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}=\frac{\mathrm{11}+\mathrm{5}\sqrt{\mathrm{5}}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}\wedge{q}=\frac{{b}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}=\frac{−\mathrm{11}+\mathrm{5}\sqrt{\mathrm{5}}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}} \\ $$$$\Rightarrow\:{pq}=\mathrm{2}^{\mathrm{8}/\mathrm{5}} \neq\mathrm{1} \\ $$

Answered by MJS_new last updated on 22/Sep/21

how to omit the ((...))^(1/5) [(∗) marks where we introduce false solutions] a^(1/5) +b^(1/5) =c^(1/5) a+5a^(4/5) b^(1/5) +10a^(3/5) b^(2/5) +10a^(2/5) b^(3/5) +5a^(1/5) b^(4/5) +b=c (∗) 5a^(1/5) b^(1/5) (a^(3/5) +2a^(2/5) b^(1/5) +2a^(1/5) b^(2/5) +b^(3/5) )=c−a−b a^(3/5) +2a^(2/5) b^(1/5) +2a^(1/5) b^(2/5) +b^(3/5) =((c−a−b)/(5a^(1/5) b^(1/5) )) (a^(1/5) +b^(1/5) )_(=c^(1/5) ) (a^(2/5) +a^(1/5) b^(1/5) +b^(2/5) )=((c−a−b)/(5a^(1/5) b^(1/5) )) a^(2/5) +a^(1/5) b^(1/5) +b^(2/5) =((c−a−b)/(5a^(1/5) b^(1/5) c^(1/5) )) now this is what I found 5(x^2 +xy+y^2 )^5 =x^(10) +y^(10) +(x+y)^(10) +3(x^5 (x+y)^5 +y^5 (x+y)^5 −x^5 y^5 ) in our case x=a^(1/5) ∧y=b^(1/5) ∧(x+y)=c^(1/5) ⇒ 5(a^(2/5) +a^(1/5) b^(1/5) +b^(2/5) )^5 =a^2 +b^2 +c^2 +3(ac+bc−ab) so we can go on (a^(2/5) +a^(1/5) b^(1/5) +b^(2/5) =((c−a−b)/(5a^(1/5) b^(1/5) c^(1/5) )))^5 (∗) (1/5)(a^2 +b^2 +c^2 +3(ac+bc−ab))=(((c−a−b)^5 )/(3125abc)) now a=x+5(√5)∧b=x−5(√5)∧c=2 −((x^2 −12x−629)/5)=((−16(x−1)^5 )/(3125(x^2 −125))) x^5 −((705)/(16))x^4 +((1915)/4)x^3 +((235545)/8)x^2 −((234355)/4)x−((49140641)/(16))=0 49140641=11^2 ×101×4021 and luckily x=11 is a solution no other solution ∈R

$$\mathrm{how}\:\mathrm{to}\:\mathrm{omit}\:\mathrm{the}\:\sqrt[{\mathrm{5}}]{...} \\ $$$$ \\ $$$$\:\:\:\:\:\left[\left(\ast\right)\:\mathrm{marks}\:\mathrm{where}\:\mathrm{we}\:\mathrm{introduce}\:\mathrm{false}\:\mathrm{solutions}\right] \\ $$$$ \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{b}^{\frac{\mathrm{1}}{\mathrm{5}}} ={c}^{\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$${a}+\mathrm{5}{a}^{\frac{\mathrm{4}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} +\mathrm{10}{a}^{\frac{\mathrm{3}}{\mathrm{5}}} {b}^{\frac{\mathrm{2}}{\mathrm{5}}} +\mathrm{10}{a}^{\frac{\mathrm{2}}{\mathrm{5}}} {b}^{\frac{\mathrm{3}}{\mathrm{5}}} +\mathrm{5}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{4}}{\mathrm{5}}} +{b}={c}\:\:\left(\ast\right) \\ $$$$\mathrm{5}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} \left({a}^{\frac{\mathrm{3}}{\mathrm{5}}} +\mathrm{2}{a}^{\frac{\mathrm{2}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} +\mathrm{2}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{2}}{\mathrm{5}}} +{b}^{\frac{\mathrm{3}}{\mathrm{5}}} \right)={c}−{a}−{b} \\ $$$${a}^{\frac{\mathrm{3}}{\mathrm{5}}} +\mathrm{2}{a}^{\frac{\mathrm{2}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} +\mathrm{2}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{2}}{\mathrm{5}}} +{b}^{\frac{\mathrm{3}}{\mathrm{5}}} =\frac{{c}−{a}−{b}}{\mathrm{5}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} } \\ $$$$\underset{={c}^{\frac{\mathrm{1}}{\mathrm{5}}} } {\underbrace{\left({a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{b}^{\frac{\mathrm{1}}{\mathrm{5}}} \right)}}\left({a}^{\frac{\mathrm{2}}{\mathrm{5}}} +{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} +{b}^{\frac{\mathrm{2}}{\mathrm{5}}} \right)=\frac{{c}−{a}−{b}}{\mathrm{5}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} } \\ $$$${a}^{\frac{\mathrm{2}}{\mathrm{5}}} +{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} +{b}^{\frac{\mathrm{2}}{\mathrm{5}}} =\frac{{c}−{a}−{b}}{\mathrm{5}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} {c}^{\frac{\mathrm{1}}{\mathrm{5}}} } \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{this}\:\mathrm{is}\:\mathrm{what}\:\mathrm{I}\:\mathrm{found} \\ $$$$\mathrm{5}\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)^{\mathrm{5}} ={x}^{\mathrm{10}} +{y}^{\mathrm{10}} +\left({x}+{y}\right)^{\mathrm{10}} +\mathrm{3}\left({x}^{\mathrm{5}} \left({x}+{y}\right)^{\mathrm{5}} +{y}^{\mathrm{5}} \left({x}+{y}\right)^{\mathrm{5}} −{x}^{\mathrm{5}} {y}^{\mathrm{5}} \right) \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:{x}={a}^{\frac{\mathrm{1}}{\mathrm{5}}} \wedge{y}={b}^{\frac{\mathrm{1}}{\mathrm{5}}} \wedge\left({x}+{y}\right)={c}^{\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$$\Rightarrow \\ $$$$\mathrm{5}\left({a}^{\frac{\mathrm{2}}{\mathrm{5}}} +{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} +{b}^{\frac{\mathrm{2}}{\mathrm{5}}} \right)^{\mathrm{5}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{3}\left({ac}+{bc}−{ab}\right) \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{can}\:\mathrm{go}\:\mathrm{on} \\ $$$$\left({a}^{\frac{\mathrm{2}}{\mathrm{5}}} +{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} +{b}^{\frac{\mathrm{2}}{\mathrm{5}}} =\frac{{c}−{a}−{b}}{\mathrm{5}{a}^{\frac{\mathrm{1}}{\mathrm{5}}} {b}^{\frac{\mathrm{1}}{\mathrm{5}}} {c}^{\frac{\mathrm{1}}{\mathrm{5}}} }\right)^{\mathrm{5}} \:\left(\ast\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{5}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{3}\left({ac}+{bc}−{ab}\right)\right)=\frac{\left({c}−{a}−{b}\right)^{\mathrm{5}} }{\mathrm{3125}{abc}} \\ $$$$\mathrm{now}\:{a}={x}+\mathrm{5}\sqrt{\mathrm{5}}\wedge{b}={x}−\mathrm{5}\sqrt{\mathrm{5}}\wedge{c}=\mathrm{2} \\ $$$$−\frac{{x}^{\mathrm{2}} −\mathrm{12}{x}−\mathrm{629}}{\mathrm{5}}=\frac{−\mathrm{16}\left({x}−\mathrm{1}\right)^{\mathrm{5}} }{\mathrm{3125}\left({x}^{\mathrm{2}} −\mathrm{125}\right)} \\ $$$${x}^{\mathrm{5}} −\frac{\mathrm{705}}{\mathrm{16}}{x}^{\mathrm{4}} +\frac{\mathrm{1915}}{\mathrm{4}}{x}^{\mathrm{3}} +\frac{\mathrm{235545}}{\mathrm{8}}{x}^{\mathrm{2}} −\frac{\mathrm{234355}}{\mathrm{4}}{x}−\frac{\mathrm{49140641}}{\mathrm{16}}=\mathrm{0} \\ $$$$\mathrm{49140641}=\mathrm{11}^{\mathrm{2}} ×\mathrm{101}×\mathrm{4021} \\ $$$$\mathrm{and}\:\mathrm{luckily}\:{x}=\mathrm{11}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{solution}\:\in\mathbb{R} \\ $$

Commented by mathdanisur last updated on 22/Sep/21

Creativ solution Ser, thank you for your attention

$$\mathrm{Creativ}\:\mathrm{solution}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$$$\mathrm{for}\:\mathrm{your}\:\mathrm{attention} \\ $$

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