Solving $ left(frac{ ( x^{3}+1 )^{3}+8 }{16}right) ^{3}+1=2x$
$begingroup$
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
edited Dec 26 '18 at 12:05
Michael Rozenberg
111k1897201
asked Dec 25 '18 at 9:25
Almot1960Almot1960
2,318825
$endgroup$
add a comment |
$begingroup$
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
edited Dec 26 '18 at 12:05
Michael Rozenberg
111k1897201
asked Dec 25 '18 at 9:25
Almot1960Almot1960
2,318825
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 26 '18 at 12:05
Michael Rozenberg
111k1897201
asked Dec 25 '18 at 9:25
Almot1960Almot1960
2,318825
$endgroup$
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
edited Dec 26 '18 at 12:05
Michael Rozenberg
111k1897201
asked Dec 25 '18 at 9:25
Almot1960Almot1960
2,318825
edited Dec 26 '18 at 12:05
Michael Rozenberg
111k1897201
asked Dec 25 '18 at 9:25
Almot1960Almot1960
2,318825
edited Dec 26 '18 at 12:05
Michael Rozenberg
111k1897201
edited Dec 26 '18 at 12:05
Michael Rozenberg
111k1897201
edited Dec 26 '18 at 12:05
Michael Rozenberg
111k1897201
111k1897201
asked Dec 25 '18 at 9:25
Almot1960Almot1960
2,318825
asked Dec 25 '18 at 9:25
Almot1960Almot1960
2,318825
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$
answered Dec 25 '18 at 11:26
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
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
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
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:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$
answered Dec 25 '18 at 11:26
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
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
add a comment |
$begingroup$
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:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$
answered Dec 25 '18 at 11:26
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
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
add a comment |
$begingroup$
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:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$
answered Dec 25 '18 at 11:26
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
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:
$$left{1,frac{-1+sqrt5}{2},frac{-1-sqrt5}{2}right}$$
answered Dec 25 '18 at 11:26
Michael RozenbergMichael Rozenberg
111k1897201
answered Dec 25 '18 at 11:26
Michael RozenbergMichael Rozenberg
111k1897201
answered Dec 25 '18 at 11:26
Michael RozenbergMichael Rozenberg
111k1897201
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
add a comment |
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
add a comment |
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$begingroup$
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