Solving $ left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}+1=2x$












2












$begingroup$



Solving :
$$left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}+1=2x$$




My Try :
$$left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}=left(dfrac{(x+1)^3+2^3}{2^4}right)^3$$



$$ x^3+a^3=(x+a)^3-3ax(x+a)$$



$$left(frac{ (x^3+3)^3-6(x^3+1)(x^3+3)}{16}right) ^{3}=left(dfrac{(x^3+3)(1-6(x^3+1))}{2^4}right)+1=2x$$



Now what ?






algebra-precalculus substitution cubic-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    The sum of 2 cubes can be factorised by $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. This might help you since there are at least 3 sum of 2 cubes there.
    $endgroup$
    – stuart stevenson
    Dec 25 '18 at 9:36






  • 2




    $begingroup$
    By inspection $x = 1$ is a solution, and consulting a CAS shows that $(textrm{l.h.s.}) - (textrm{r.h.s.}) = (x - 1) (x^2 + x - 1) p(x)$ for some irreducible polynomial $p$ of degree $24$ with no real roots.
    $endgroup$
    – Travis
    Dec 25 '18 at 9:50






  • 1




    $begingroup$
    Perhaps there is some slick trick to solve this, but as it appears this seems to be a horrible, almost criminal, equation to solve.
    $endgroup$
    – DonAntonio
    Dec 25 '18 at 10:29
















2












$begingroup$



Solving :
$$left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}+1=2x$$




My Try :
$$left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}=left(dfrac{(x+1)^3+2^3}{2^4}right)^3$$



$$ x^3+a^3=(x+a)^3-3ax(x+a)$$



$$left(frac{ (x^3+3)^3-6(x^3+1)(x^3+3)}{16}right) ^{3}=left(dfrac{(x^3+3)(1-6(x^3+1))}{2^4}right)+1=2x$$



Now what ?






algebra-precalculus substitution cubic-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    The sum of 2 cubes can be factorised by $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. This might help you since there are at least 3 sum of 2 cubes there.
    $endgroup$
    – stuart stevenson
    Dec 25 '18 at 9:36






  • 2




    $begingroup$
    By inspection $x = 1$ is a solution, and consulting a CAS shows that $(textrm{l.h.s.}) - (textrm{r.h.s.}) = (x - 1) (x^2 + x - 1) p(x)$ for some irreducible polynomial $p$ of degree $24$ with no real roots.
    $endgroup$
    – Travis
    Dec 25 '18 at 9:50






  • 1




    $begingroup$
    Perhaps there is some slick trick to solve this, but as it appears this seems to be a horrible, almost criminal, equation to solve.
    $endgroup$
    – DonAntonio
    Dec 25 '18 at 10:29














2












2








2


1



$begingroup$



Solving :
$$left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}+1=2x$$




My Try :
$$left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}=left(dfrac{(x+1)^3+2^3}{2^4}right)^3$$



$$ x^3+a^3=(x+a)^3-3ax(x+a)$$



$$left(frac{ (x^3+3)^3-6(x^3+1)(x^3+3)}{16}right) ^{3}=left(dfrac{(x^3+3)(1-6(x^3+1))}{2^4}right)+1=2x$$



Now what ?






algebra-precalculus substitution cubic-equations







share|cite|improve this question











$endgroup$





Solving :
$$left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}+1=2x$$




My Try :
$$left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}=left(dfrac{(x+1)^3+2^3}{2^4}right)^3$$



$$ x^3+a^3=(x+a)^3-3ax(x+a)$$



$$left(frac{ (x^3+3)^3-6(x^3+1)(x^3+3)}{16}right) ^{3}=left(dfrac{(x^3+3)(1-6(x^3+1))}{2^4}right)+1=2x$$



Now what ?






algebra-precalculus substitution cubic-equations




algebra-precalculus substitution cubic-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 12:05









Michael Rozenberg

111k1897201




111k1897201










asked Dec 25 '18 at 9:25









Almot1960Almot1960

2,318825




2,318825












  • $begingroup$
    The sum of 2 cubes can be factorised by $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. This might help you since there are at least 3 sum of 2 cubes there.
    $endgroup$
    – stuart stevenson
    Dec 25 '18 at 9:36






  • 2




    $begingroup$
    By inspection $x = 1$ is a solution, and consulting a CAS shows that $(textrm{l.h.s.}) - (textrm{r.h.s.}) = (x - 1) (x^2 + x - 1) p(x)$ for some irreducible polynomial $p$ of degree $24$ with no real roots.
    $endgroup$
    – Travis
    Dec 25 '18 at 9:50






  • 1




    $begingroup$
    Perhaps there is some slick trick to solve this, but as it appears this seems to be a horrible, almost criminal, equation to solve.
    $endgroup$
    – DonAntonio
    Dec 25 '18 at 10:29


















  • $begingroup$
    The sum of 2 cubes can be factorised by $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. This might help you since there are at least 3 sum of 2 cubes there.
    $endgroup$
    – stuart stevenson
    Dec 25 '18 at 9:36






  • 2




    $begingroup$
    By inspection $x = 1$ is a solution, and consulting a CAS shows that $(textrm{l.h.s.}) - (textrm{r.h.s.}) = (x - 1) (x^2 + x - 1) p(x)$ for some irreducible polynomial $p$ of degree $24$ with no real roots.
    $endgroup$
    – Travis
    Dec 25 '18 at 9:50






  • 1




    $begingroup$
    Perhaps there is some slick trick to solve this, but as it appears this seems to be a horrible, almost criminal, equation to solve.
    $endgroup$
    – DonAntonio
    Dec 25 '18 at 10:29
















$begingroup$
The sum of 2 cubes can be factorised by $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. This might help you since there are at least 3 sum of 2 cubes there.
$endgroup$
– stuart stevenson
Dec 25 '18 at 9:36




$begingroup$
The sum of 2 cubes can be factorised by $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. This might help you since there are at least 3 sum of 2 cubes there.
$endgroup$
– stuart stevenson
Dec 25 '18 at 9:36




2




2




$begingroup$
By inspection $x = 1$ is a solution, and consulting a CAS shows that $(textrm{l.h.s.}) - (textrm{r.h.s.}) = (x - 1) (x^2 + x - 1) p(x)$ for some irreducible polynomial $p$ of degree $24$ with no real roots.
$endgroup$
– Travis
Dec 25 '18 at 9:50




$begingroup$
By inspection $x = 1$ is a solution, and consulting a CAS shows that $(textrm{l.h.s.}) - (textrm{r.h.s.}) = (x - 1) (x^2 + x - 1) p(x)$ for some irreducible polynomial $p$ of degree $24$ with no real roots.
$endgroup$
– Travis
Dec 25 '18 at 9:50




1




1




$begingroup$
Perhaps there is some slick trick to solve this, but as it appears this seems to be a horrible, almost criminal, equation to solve.
$endgroup$
– DonAntonio
Dec 25 '18 at 10:29




$begingroup$
Perhaps there is some slick trick to solve this, but as it appears this seems to be a horrible, almost criminal, equation to solve.
$endgroup$
– DonAntonio
Dec 25 '18 at 10:29










1 Answer
1






active

oldest

votes


















6












$begingroup$

Let $frac{x^3+1}{2}=y$ and $sqrt[3]{2x-1}=z.$



Thus, from the given we obtain $frac{y^3+1}{2}=z.$



Now, let $x>y$.



Thus, $$x>y=frac{x^3+1}{2}>frac{y^3+1}{2}=z,$$ which says that
$$x>y>z.$$
But from $y^3+1=2z$ and $z^3+1=2x$ we obtain
$$0=y^3-z^3+2(x-z)>0,$$ which is a contradiction.



By the same way we'll get a contradiction for $x<y$.



Id est $x=y$, $$x^3-2x+1=0$$ or
$$x^3-x^2+x^2-x-x+1=0$$ or
$$(x-1)(x^2+x-1)=0,$$
which gives the answer:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Slight variant: define $fcolonmathbb{R}tomathbb{R}$, $xmapstofrac{x^3+1}{2}$. $f$ is strictly increasing on $mathbb{R}$, so if $x lessgtr f(x)$, then $f(x) lessgtr f(f(x))$, therefore $f(f(x)) lessgtr f(f(f(x)))$, therefore $x lessgtr f(f(f(x)))$. Hence, $x = f(f(f(x)))$ iff $x = f(x)$.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 14:07










  • $begingroup$
    @Calum Gilhooley I saw this way. I just wrote this idea differently.
    $endgroup$
    – Michael Rozenberg
    Dec 25 '18 at 19:03






  • 1




    $begingroup$
    It was merely my take on what you had written, only a "slight" variant, as I said. Yours is a nice answer; I just wanted to make it a little easier to see at a glance.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 20:06












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051964%2fsolving-left-frac-x31-38-16-right-31-2x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Let $frac{x^3+1}{2}=y$ and $sqrt[3]{2x-1}=z.$



Thus, from the given we obtain $frac{y^3+1}{2}=z.$



Now, let $x>y$.



Thus, $$x>y=frac{x^3+1}{2}>frac{y^3+1}{2}=z,$$ which says that
$$x>y>z.$$
But from $y^3+1=2z$ and $z^3+1=2x$ we obtain
$$0=y^3-z^3+2(x-z)>0,$$ which is a contradiction.



By the same way we'll get a contradiction for $x<y$.



Id est $x=y$, $$x^3-2x+1=0$$ or
$$x^3-x^2+x^2-x-x+1=0$$ or
$$(x-1)(x^2+x-1)=0,$$
which gives the answer:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Slight variant: define $fcolonmathbb{R}tomathbb{R}$, $xmapstofrac{x^3+1}{2}$. $f$ is strictly increasing on $mathbb{R}$, so if $x lessgtr f(x)$, then $f(x) lessgtr f(f(x))$, therefore $f(f(x)) lessgtr f(f(f(x)))$, therefore $x lessgtr f(f(f(x)))$. Hence, $x = f(f(f(x)))$ iff $x = f(x)$.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 14:07










  • $begingroup$
    @Calum Gilhooley I saw this way. I just wrote this idea differently.
    $endgroup$
    – Michael Rozenberg
    Dec 25 '18 at 19:03






  • 1




    $begingroup$
    It was merely my take on what you had written, only a "slight" variant, as I said. Yours is a nice answer; I just wanted to make it a little easier to see at a glance.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 20:06
















6












$begingroup$

Let $frac{x^3+1}{2}=y$ and $sqrt[3]{2x-1}=z.$



Thus, from the given we obtain $frac{y^3+1}{2}=z.$



Now, let $x>y$.



Thus, $$x>y=frac{x^3+1}{2}>frac{y^3+1}{2}=z,$$ which says that
$$x>y>z.$$
But from $y^3+1=2z$ and $z^3+1=2x$ we obtain
$$0=y^3-z^3+2(x-z)>0,$$ which is a contradiction.



By the same way we'll get a contradiction for $x<y$.



Id est $x=y$, $$x^3-2x+1=0$$ or
$$x^3-x^2+x^2-x-x+1=0$$ or
$$(x-1)(x^2+x-1)=0,$$
which gives the answer:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Slight variant: define $fcolonmathbb{R}tomathbb{R}$, $xmapstofrac{x^3+1}{2}$. $f$ is strictly increasing on $mathbb{R}$, so if $x lessgtr f(x)$, then $f(x) lessgtr f(f(x))$, therefore $f(f(x)) lessgtr f(f(f(x)))$, therefore $x lessgtr f(f(f(x)))$. Hence, $x = f(f(f(x)))$ iff $x = f(x)$.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 14:07










  • $begingroup$
    @Calum Gilhooley I saw this way. I just wrote this idea differently.
    $endgroup$
    – Michael Rozenberg
    Dec 25 '18 at 19:03






  • 1




    $begingroup$
    It was merely my take on what you had written, only a "slight" variant, as I said. Yours is a nice answer; I just wanted to make it a little easier to see at a glance.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 20:06














6












6








6





$begingroup$

Let $frac{x^3+1}{2}=y$ and $sqrt[3]{2x-1}=z.$



Thus, from the given we obtain $frac{y^3+1}{2}=z.$



Now, let $x>y$.



Thus, $$x>y=frac{x^3+1}{2}>frac{y^3+1}{2}=z,$$ which says that
$$x>y>z.$$
But from $y^3+1=2z$ and $z^3+1=2x$ we obtain
$$0=y^3-z^3+2(x-z)>0,$$ which is a contradiction.



By the same way we'll get a contradiction for $x<y$.



Id est $x=y$, $$x^3-2x+1=0$$ or
$$x^3-x^2+x^2-x-x+1=0$$ or
$$(x-1)(x^2+x-1)=0,$$
which gives the answer:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$






share|cite|improve this answer









$endgroup$



Let $frac{x^3+1}{2}=y$ and $sqrt[3]{2x-1}=z.$



Thus, from the given we obtain $frac{y^3+1}{2}=z.$



Now, let $x>y$.



Thus, $$x>y=frac{x^3+1}{2}>frac{y^3+1}{2}=z,$$ which says that
$$x>y>z.$$
But from $y^3+1=2z$ and $z^3+1=2x$ we obtain
$$0=y^3-z^3+2(x-z)>0,$$ which is a contradiction.



By the same way we'll get a contradiction for $x<y$.



Id est $x=y$, $$x^3-2x+1=0$$ or
$$x^3-x^2+x^2-x-x+1=0$$ or
$$(x-1)(x^2+x-1)=0,$$
which gives the answer:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 11:26









Michael RozenbergMichael Rozenberg

111k1897201




111k1897201








  • 1




    $begingroup$
    Slight variant: define $fcolonmathbb{R}tomathbb{R}$, $xmapstofrac{x^3+1}{2}$. $f$ is strictly increasing on $mathbb{R}$, so if $x lessgtr f(x)$, then $f(x) lessgtr f(f(x))$, therefore $f(f(x)) lessgtr f(f(f(x)))$, therefore $x lessgtr f(f(f(x)))$. Hence, $x = f(f(f(x)))$ iff $x = f(x)$.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 14:07










  • $begingroup$
    @Calum Gilhooley I saw this way. I just wrote this idea differently.
    $endgroup$
    – Michael Rozenberg
    Dec 25 '18 at 19:03






  • 1




    $begingroup$
    It was merely my take on what you had written, only a "slight" variant, as I said. Yours is a nice answer; I just wanted to make it a little easier to see at a glance.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 20:06














  • 1




    $begingroup$
    Slight variant: define $fcolonmathbb{R}tomathbb{R}$, $xmapstofrac{x^3+1}{2}$. $f$ is strictly increasing on $mathbb{R}$, so if $x lessgtr f(x)$, then $f(x) lessgtr f(f(x))$, therefore $f(f(x)) lessgtr f(f(f(x)))$, therefore $x lessgtr f(f(f(x)))$. Hence, $x = f(f(f(x)))$ iff $x = f(x)$.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 14:07










  • $begingroup$
    @Calum Gilhooley I saw this way. I just wrote this idea differently.
    $endgroup$
    – Michael Rozenberg
    Dec 25 '18 at 19:03






  • 1




    $begingroup$
    It was merely my take on what you had written, only a "slight" variant, as I said. Yours is a nice answer; I just wanted to make it a little easier to see at a glance.
    $endgroup$
    – Calum Gilhooley
    Dec 25 '18 at 20:06








1




1




$begingroup$
Slight variant: define $fcolonmathbb{R}tomathbb{R}$, $xmapstofrac{x^3+1}{2}$. $f$ is strictly increasing on $mathbb{R}$, so if $x lessgtr f(x)$, then $f(x) lessgtr f(f(x))$, therefore $f(f(x)) lessgtr f(f(f(x)))$, therefore $x lessgtr f(f(f(x)))$. Hence, $x = f(f(f(x)))$ iff $x = f(x)$.
$endgroup$
– Calum Gilhooley
Dec 25 '18 at 14:07




$begingroup$
Slight variant: define $fcolonmathbb{R}tomathbb{R}$, $xmapstofrac{x^3+1}{2}$. $f$ is strictly increasing on $mathbb{R}$, so if $x lessgtr f(x)$, then $f(x) lessgtr f(f(x))$, therefore $f(f(x)) lessgtr f(f(f(x)))$, therefore $x lessgtr f(f(f(x)))$. Hence, $x = f(f(f(x)))$ iff $x = f(x)$.
$endgroup$
– Calum Gilhooley
Dec 25 '18 at 14:07












$begingroup$
@Calum Gilhooley I saw this way. I just wrote this idea differently.
$endgroup$
– Michael Rozenberg
Dec 25 '18 at 19:03




$begingroup$
@Calum Gilhooley I saw this way. I just wrote this idea differently.
$endgroup$
– Michael Rozenberg
Dec 25 '18 at 19:03




1




1




$begingroup$
It was merely my take on what you had written, only a "slight" variant, as I said. Yours is a nice answer; I just wanted to make it a little easier to see at a glance.
$endgroup$
– Calum Gilhooley
Dec 25 '18 at 20:06




$begingroup$
It was merely my take on what you had written, only a "slight" variant, as I said. Yours is a nice answer; I just wanted to make it a little easier to see at a glance.
$endgroup$
– Calum Gilhooley
Dec 25 '18 at 20:06


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051964%2fsolving-left-frac-x31-38-16-right-31-2x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more