To show that 2 roots of $(x-a)^3$ + $(x-b)^3$ +$(x-c)^3 = 0 $ are complex
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I am afraid that this question might be marked duplicate, but I simply had to ask if there is a way to show that the equation $(x-a)^3$ + $(x-b)^3$ +$(x-c)^3 = 0$ has only 1 real root? One of my teachers in high school posed the problem, and here's the twist, asked for a solution that does NOT invoke Rolle's theorem. He gave a hint that we could use the graph of said equation, but I don't see how that helps. I tried using Descartes' rules and Vieta's formulae, but they don't seem to lead anywhere. Thanks for helping!
EDIT:
After some insights from gracious responders,I see that my question boils down to this:Can we construct some artfully watertight piece of argument which shows that an increasing function has one and only one real root?
algebra-precalculus
asked Sep 8 at 11:18
abhibrata ganguly
137
add a comment |Â
up vote
2
down vote
favorite
I am afraid that this question might be marked duplicate, but I simply had to ask if there is a way to show that the equation $(x-a)^3$ + $(x-b)^3$ +$(x-c)^3 = 0$ has only 1 real root? One of my teachers in high school posed the problem, and here's the twist, asked for a solution that does NOT invoke Rolle's theorem. He gave a hint that we could use the graph of said equation, but I don't see how that helps. I tried using Descartes' rules and Vieta's formulae, but they don't seem to lead anywhere. Thanks for helping!
EDIT:
After some insights from gracious responders,I see that my question boils down to this:Can we construct some artfully watertight piece of argument which shows that an increasing function has one and only one real root?
algebra-precalculus
asked Sep 8 at 11:18
abhibrata ganguly
137
5
It's an increasing function....
– Lord Shark the Unknown
Sep 8 at 11:22
1
Suppose $a$, $b$ and $c$ are real. $(x-a)^3+(x-b)^3+(x-c)^3$ is straightly increasing, so its graph cuts the $x$-axis once only.
– Yuta
Sep 8 at 11:22
Thanks for helping. But can we show that it cuts the x-axis only once?By algebra.I can see that it follows the graph of x^3,But that is not a proof.
– abhibrata ganguly
Sep 8 at 11:27
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am afraid that this question might be marked duplicate, but I simply had to ask if there is a way to show that the equation $(x-a)^3$ + $(x-b)^3$ +$(x-c)^3 = 0$ has only 1 real root? One of my teachers in high school posed the problem, and here's the twist, asked for a solution that does NOT invoke Rolle's theorem. He gave a hint that we could use the graph of said equation, but I don't see how that helps. I tried using Descartes' rules and Vieta's formulae, but they don't seem to lead anywhere. Thanks for helping!
EDIT:
After some insights from gracious responders,I see that my question boils down to this:Can we construct some artfully watertight piece of argument which shows that an increasing function has one and only one real root?
algebra-precalculus
asked Sep 8 at 11:18
abhibrata ganguly
137
I am afraid that this question might be marked duplicate, but I simply had to ask if there is a way to show that the equation $(x-a)^3$ + $(x-b)^3$ +$(x-c)^3 = 0$ has only 1 real root? One of my teachers in high school posed the problem, and here's the twist, asked for a solution that does NOT invoke Rolle's theorem. He gave a hint that we could use the graph of said equation, but I don't see how that helps. I tried using Descartes' rules and Vieta's formulae, but they don't seem to lead anywhere. Thanks for helping!
EDIT:
After some insights from gracious responders,I see that my question boils down to this:Can we construct some artfully watertight piece of argument which shows that an increasing function has one and only one real root?
algebra-precalculus
algebra-precalculus
asked Sep 8 at 11:18
abhibrata ganguly
137
asked Sep 8 at 11:18
abhibrata ganguly
137
edited Sep 8 at 11:56
asked Sep 8 at 11:18
abhibrata ganguly
137
asked Sep 8 at 11:18
abhibrata ganguly
137
asked Sep 8 at 11:18
abhibrata ganguly
137
137
5
It's an increasing function....
– Lord Shark the Unknown
Sep 8 at 11:22
1
Suppose $a$, $b$ and $c$ are real. $(x-a)^3+(x-b)^3+(x-c)^3$ is straightly increasing, so its graph cuts the $x$-axis once only.
– Yuta
Sep 8 at 11:22
Thanks for helping. But can we show that it cuts the x-axis only once?By algebra.I can see that it follows the graph of x^3,But that is not a proof.
– abhibrata ganguly
Sep 8 at 11:27
add a comment |Â
5
It's an increasing function....
– Lord Shark the Unknown
Sep 8 at 11:22
1
Suppose $a$, $b$ and $c$ are real. $(x-a)^3+(x-b)^3+(x-c)^3$ is straightly increasing, so its graph cuts the $x$-axis once only.
– Yuta
Sep 8 at 11:22
Thanks for helping. But can we show that it cuts the x-axis only once?By algebra.I can see that it follows the graph of x^3,But that is not a proof.
– abhibrata ganguly
Sep 8 at 11:27
5
5
It's an increasing function....
– Lord Shark the Unknown
Sep 8 at 11:22
It's an increasing function....
– Lord Shark the Unknown
Sep 8 at 11:22
1
1
Suppose $a$, $b$ and $c$ are real. $(x-a)^3+(x-b)^3+(x-c)^3$ is straightly increasing, so its graph cuts the $x$-axis once only.
– Yuta
Sep 8 at 11:22
Suppose $a$, $b$ and $c$ are real. $(x-a)^3+(x-b)^3+(x-c)^3$ is straightly increasing, so its graph cuts the $x$-axis once only.
– Yuta
Sep 8 at 11:22
Thanks for helping. But can we show that it cuts the x-axis only once?By algebra.I can see that it follows the graph of x^3,But that is not a proof.
– abhibrata ganguly
Sep 8 at 11:27
Thanks for helping. But can we show that it cuts the x-axis only once?By algebra.I can see that it follows the graph of x^3,But that is not a proof.
– abhibrata ganguly
Sep 8 at 11:27
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
As the comments already suggest, calculate $f'(x)$,
$$f'(x)=3left((x-a)^2+(x-b)^2+(x-c)^2right)$$
Can you state something about $f'(x)$ that will help you determine it has only one root?
answered Sep 8 at 11:26
prog_SAHIL
1,527318
This equation forces x=a=b=c, at the maxima or minima betwwen three roots,thus causing degeneracy, when we don't have any assumption regarding a,b,c except they are reals.
– abhibrata ganguly
Sep 8 at 11:29
None of the comments suggested differentiating $f$.
– Lord Shark the Unknown
Sep 8 at 11:30
@abhibrataganguly, It isn't necessary that a function has a maxima or a minima.
– prog_SAHIL
Sep 8 at 11:31
@LordSharktheUnknown, I am just proving it is an increasing function. (The comments were suggesting that). This seems like the easiest way.
– prog_SAHIL
Sep 8 at 11:32
It is easy to visualize.I guess that's why he forbid Rolle's theorem.
– abhibrata ganguly
Sep 8 at 11:33
 |Â
show 4 more comments
up vote
1
down vote
Hint: Consider the equation
$$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)=0$$
Then defining
$$f(x)=3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)$$ and use Calculus.
answered Sep 8 at 11:22
Dr. Sonnhard Graubner
69k32761
I think there are too many cases to be dealt with here if you want to use Descartes's Rule of Signs. The only coefficient that we know the sign is the coefficient of $x$.
– Batominovski
Sep 8 at 11:29
Thank you for your hint,just corrected.
– Dr. Sonnhard Graubner
Sep 8 at 11:30
Without calculus?
– abhibrata ganguly
Sep 8 at 11:33
Then you can use laws of Cardano's formula about the discriminant.
– Dr. Sonnhard Graubner
Sep 8 at 11:37
Would you please write the discriminant of your cubic and add it in your answer ?
– Claude Leibovici
Sep 8 at 14:20
add a comment |Â
up vote
1
down vote
Hint: If $f(x)$ has $r$ real roots, then $f'(x)$ will have atleast $r-1$ real roots.
Then you just have to prove that $f'(x)$ has no real roots.
answered Sep 8 at 11:36
paulplusx
1,056318
1
@prog_SAHIL Please read my statement again. I said if $f(x)$ has $r$ real roots.
– paulplusx
Sep 8 at 11:40
I misread. Deleted.
– prog_SAHIL
Sep 8 at 11:46
How do I proceed?
– abhibrata ganguly
Sep 8 at 11:52
@abhibrataganguly Can you prove $f'(x)$ has no real roots?
– paulplusx
Sep 8 at 11:56
Yup, x=a=b=c,which can't be allowed as I said in an earlier response.But is there a non-calculus way of showing the idea?
– abhibrata ganguly
Sep 8 at 11:58
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
As the comments already suggest, calculate $f'(x)$,
$$f'(x)=3left((x-a)^2+(x-b)^2+(x-c)^2right)$$
Can you state something about $f'(x)$ that will help you determine it has only one root?
answered Sep 8 at 11:26
prog_SAHIL
1,527318
This equation forces x=a=b=c, at the maxima or minima betwwen three roots,thus causing degeneracy, when we don't have any assumption regarding a,b,c except they are reals.
– abhibrata ganguly
Sep 8 at 11:29
None of the comments suggested differentiating $f$.
– Lord Shark the Unknown
Sep 8 at 11:30
@abhibrataganguly, It isn't necessary that a function has a maxima or a minima.
– prog_SAHIL
Sep 8 at 11:31
@LordSharktheUnknown, I am just proving it is an increasing function. (The comments were suggesting that). This seems like the easiest way.
– prog_SAHIL
Sep 8 at 11:32
It is easy to visualize.I guess that's why he forbid Rolle's theorem.
– abhibrata ganguly
Sep 8 at 11:33
 |Â
show 4 more comments
up vote
2
down vote
As the comments already suggest, calculate $f'(x)$,
$$f'(x)=3left((x-a)^2+(x-b)^2+(x-c)^2right)$$
Can you state something about $f'(x)$ that will help you determine it has only one root?
answered Sep 8 at 11:26
prog_SAHIL
1,527318
This equation forces x=a=b=c, at the maxima or minima betwwen three roots,thus causing degeneracy, when we don't have any assumption regarding a,b,c except they are reals.
– abhibrata ganguly
Sep 8 at 11:29
None of the comments suggested differentiating $f$.
– Lord Shark the Unknown
Sep 8 at 11:30
@abhibrataganguly, It isn't necessary that a function has a maxima or a minima.
– prog_SAHIL
Sep 8 at 11:31
@LordSharktheUnknown, I am just proving it is an increasing function. (The comments were suggesting that). This seems like the easiest way.
– prog_SAHIL
Sep 8 at 11:32
It is easy to visualize.I guess that's why he forbid Rolle's theorem.
– abhibrata ganguly
Sep 8 at 11:33
 |Â
show 4 more comments
up vote
2
down vote
up vote
2
down vote
As the comments already suggest, calculate $f'(x)$,
$$f'(x)=3left((x-a)^2+(x-b)^2+(x-c)^2right)$$
Can you state something about $f'(x)$ that will help you determine it has only one root?
answered Sep 8 at 11:26
prog_SAHIL
1,527318
As the comments already suggest, calculate $f'(x)$,
$$f'(x)=3left((x-a)^2+(x-b)^2+(x-c)^2right)$$
Can you state something about $f'(x)$ that will help you determine it has only one root?
answered Sep 8 at 11:26
prog_SAHIL
1,527318
answered Sep 8 at 11:26
prog_SAHIL
1,527318
answered Sep 8 at 11:26
prog_SAHIL
1,527318
answered Sep 8 at 11:26
prog_SAHIL
1,527318
1,527318
This equation forces x=a=b=c, at the maxima or minima betwwen three roots,thus causing degeneracy, when we don't have any assumption regarding a,b,c except they are reals.
– abhibrata ganguly
Sep 8 at 11:29
None of the comments suggested differentiating $f$.
– Lord Shark the Unknown
Sep 8 at 11:30
@abhibrataganguly, It isn't necessary that a function has a maxima or a minima.
– prog_SAHIL
Sep 8 at 11:31
@LordSharktheUnknown, I am just proving it is an increasing function. (The comments were suggesting that). This seems like the easiest way.
– prog_SAHIL
Sep 8 at 11:32
It is easy to visualize.I guess that's why he forbid Rolle's theorem.
– abhibrata ganguly
Sep 8 at 11:33
 |Â
show 4 more comments
This equation forces x=a=b=c, at the maxima or minima betwwen three roots,thus causing degeneracy, when we don't have any assumption regarding a,b,c except they are reals.
– abhibrata ganguly
Sep 8 at 11:29
None of the comments suggested differentiating $f$.
– Lord Shark the Unknown
Sep 8 at 11:30
@abhibrataganguly, It isn't necessary that a function has a maxima or a minima.
– prog_SAHIL
Sep 8 at 11:31
@LordSharktheUnknown, I am just proving it is an increasing function. (The comments were suggesting that). This seems like the easiest way.
– prog_SAHIL
Sep 8 at 11:32
It is easy to visualize.I guess that's why he forbid Rolle's theorem.
– abhibrata ganguly
Sep 8 at 11:33
This equation forces x=a=b=c, at the maxima or minima betwwen three roots,thus causing degeneracy, when we don't have any assumption regarding a,b,c except they are reals.
– abhibrata ganguly
Sep 8 at 11:29
This equation forces x=a=b=c, at the maxima or minima betwwen three roots,thus causing degeneracy, when we don't have any assumption regarding a,b,c except they are reals.
– abhibrata ganguly
Sep 8 at 11:29
None of the comments suggested differentiating $f$.
– Lord Shark the Unknown
Sep 8 at 11:30
None of the comments suggested differentiating $f$.
– Lord Shark the Unknown
Sep 8 at 11:30
@abhibrataganguly, It isn't necessary that a function has a maxima or a minima.
– prog_SAHIL
Sep 8 at 11:31
@abhibrataganguly, It isn't necessary that a function has a maxima or a minima.
– prog_SAHIL
Sep 8 at 11:31
@LordSharktheUnknown, I am just proving it is an increasing function. (The comments were suggesting that). This seems like the easiest way.
– prog_SAHIL
Sep 8 at 11:32
@LordSharktheUnknown, I am just proving it is an increasing function. (The comments were suggesting that). This seems like the easiest way.
– prog_SAHIL
Sep 8 at 11:32
It is easy to visualize.I guess that's why he forbid Rolle's theorem.
– abhibrata ganguly
Sep 8 at 11:33
It is easy to visualize.I guess that's why he forbid Rolle's theorem.
– abhibrata ganguly
Sep 8 at 11:33
 |Â
show 4 more comments
up vote
1
down vote
Hint: Consider the equation
$$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)=0$$
Then defining
$$f(x)=3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)$$ and use Calculus.
answered Sep 8 at 11:22
Dr. Sonnhard Graubner
69k32761
I think there are too many cases to be dealt with here if you want to use Descartes's Rule of Signs. The only coefficient that we know the sign is the coefficient of $x$.
– Batominovski
Sep 8 at 11:29
Thank you for your hint,just corrected.
– Dr. Sonnhard Graubner
Sep 8 at 11:30
Without calculus?
– abhibrata ganguly
Sep 8 at 11:33
Then you can use laws of Cardano's formula about the discriminant.
– Dr. Sonnhard Graubner
Sep 8 at 11:37
Would you please write the discriminant of your cubic and add it in your answer ?
– Claude Leibovici
Sep 8 at 14:20
add a comment |Â
up vote
1
down vote
Hint: Consider the equation
$$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)=0$$
Then defining
$$f(x)=3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)$$ and use Calculus.
answered Sep 8 at 11:22
Dr. Sonnhard Graubner
69k32761
I think there are too many cases to be dealt with here if you want to use Descartes's Rule of Signs. The only coefficient that we know the sign is the coefficient of $x$.
– Batominovski
Sep 8 at 11:29
Thank you for your hint,just corrected.
– Dr. Sonnhard Graubner
Sep 8 at 11:30
Without calculus?
– abhibrata ganguly
Sep 8 at 11:33
Then you can use laws of Cardano's formula about the discriminant.
– Dr. Sonnhard Graubner
Sep 8 at 11:37
Would you please write the discriminant of your cubic and add it in your answer ?
– Claude Leibovici
Sep 8 at 14:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Consider the equation
$$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)=0$$
Then defining
$$f(x)=3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)$$ and use Calculus.
answered Sep 8 at 11:22
Dr. Sonnhard Graubner
69k32761
Hint: Consider the equation
$$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)=0$$
Then defining
$$f(x)=3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-(a^3+b^3+c^3)$$ and use Calculus.
answered Sep 8 at 11:22
Dr. Sonnhard Graubner
69k32761
edited Sep 8 at 11:33
answered Sep 8 at 11:22
Dr. Sonnhard Graubner
69k32761
answered Sep 8 at 11:22
Dr. Sonnhard Graubner
69k32761
answered Sep 8 at 11:22
Dr. Sonnhard Graubner
69k32761
69k32761
I think there are too many cases to be dealt with here if you want to use Descartes's Rule of Signs. The only coefficient that we know the sign is the coefficient of $x$.
– Batominovski
Sep 8 at 11:29
Thank you for your hint,just corrected.
– Dr. Sonnhard Graubner
Sep 8 at 11:30
Without calculus?
– abhibrata ganguly
Sep 8 at 11:33
Then you can use laws of Cardano's formula about the discriminant.
– Dr. Sonnhard Graubner
Sep 8 at 11:37
Would you please write the discriminant of your cubic and add it in your answer ?
– Claude Leibovici
Sep 8 at 14:20
add a comment |Â
I think there are too many cases to be dealt with here if you want to use Descartes's Rule of Signs. The only coefficient that we know the sign is the coefficient of $x$.
– Batominovski
Sep 8 at 11:29
Thank you for your hint,just corrected.
– Dr. Sonnhard Graubner
Sep 8 at 11:30
Without calculus?
– abhibrata ganguly
Sep 8 at 11:33
Then you can use laws of Cardano's formula about the discriminant.
– Dr. Sonnhard Graubner
Sep 8 at 11:37
Would you please write the discriminant of your cubic and add it in your answer ?
– Claude Leibovici
Sep 8 at 14:20
I think there are too many cases to be dealt with here if you want to use Descartes's Rule of Signs. The only coefficient that we know the sign is the coefficient of $x$.
– Batominovski
Sep 8 at 11:29
I think there are too many cases to be dealt with here if you want to use Descartes's Rule of Signs. The only coefficient that we know the sign is the coefficient of $x$.
– Batominovski
Sep 8 at 11:29
Thank you for your hint,just corrected.
– Dr. Sonnhard Graubner
Sep 8 at 11:30
Thank you for your hint,just corrected.
– Dr. Sonnhard Graubner
Sep 8 at 11:30
Without calculus?
– abhibrata ganguly
Sep 8 at 11:33
Without calculus?
– abhibrata ganguly
Sep 8 at 11:33
Then you can use laws of Cardano's formula about the discriminant.
– Dr. Sonnhard Graubner
Sep 8 at 11:37
Then you can use laws of Cardano's formula about the discriminant.
– Dr. Sonnhard Graubner
Sep 8 at 11:37
Would you please write the discriminant of your cubic and add it in your answer ?
– Claude Leibovici
Sep 8 at 14:20
Would you please write the discriminant of your cubic and add it in your answer ?
– Claude Leibovici
Sep 8 at 14:20
add a comment |Â
up vote
1
down vote
Hint: If $f(x)$ has $r$ real roots, then $f'(x)$ will have atleast $r-1$ real roots.
Then you just have to prove that $f'(x)$ has no real roots.
answered Sep 8 at 11:36
paulplusx
1,056318
1
@prog_SAHIL Please read my statement again. I said if $f(x)$ has $r$ real roots.
– paulplusx
Sep 8 at 11:40
I misread. Deleted.
– prog_SAHIL
Sep 8 at 11:46
How do I proceed?
– abhibrata ganguly
Sep 8 at 11:52
@abhibrataganguly Can you prove $f'(x)$ has no real roots?
– paulplusx
Sep 8 at 11:56
Yup, x=a=b=c,which can't be allowed as I said in an earlier response.But is there a non-calculus way of showing the idea?
– abhibrata ganguly
Sep 8 at 11:58
 |Â
show 1 more comment
up vote
1
down vote
Hint: If $f(x)$ has $r$ real roots, then $f'(x)$ will have atleast $r-1$ real roots.
Then you just have to prove that $f'(x)$ has no real roots.
answered Sep 8 at 11:36
paulplusx
1,056318
1
@prog_SAHIL Please read my statement again. I said if $f(x)$ has $r$ real roots.
– paulplusx
Sep 8 at 11:40
I misread. Deleted.
– prog_SAHIL
Sep 8 at 11:46
How do I proceed?
– abhibrata ganguly
Sep 8 at 11:52
@abhibrataganguly Can you prove $f'(x)$ has no real roots?
– paulplusx
Sep 8 at 11:56
Yup, x=a=b=c,which can't be allowed as I said in an earlier response.But is there a non-calculus way of showing the idea?
– abhibrata ganguly
Sep 8 at 11:58
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
Hint: If $f(x)$ has $r$ real roots, then $f'(x)$ will have atleast $r-1$ real roots.
Then you just have to prove that $f'(x)$ has no real roots.
answered Sep 8 at 11:36
paulplusx
1,056318
Hint: If $f(x)$ has $r$ real roots, then $f'(x)$ will have atleast $r-1$ real roots.
Then you just have to prove that $f'(x)$ has no real roots.
answered Sep 8 at 11:36
paulplusx
1,056318
edited Sep 8 at 12:07
answered Sep 8 at 11:36
paulplusx
1,056318
answered Sep 8 at 11:36
paulplusx
1,056318
answered Sep 8 at 11:36
paulplusx
1,056318
1,056318
1
@prog_SAHIL Please read my statement again. I said if $f(x)$ has $r$ real roots.
– paulplusx
Sep 8 at 11:40
I misread. Deleted.
– prog_SAHIL
Sep 8 at 11:46
How do I proceed?
– abhibrata ganguly
Sep 8 at 11:52
@abhibrataganguly Can you prove $f'(x)$ has no real roots?
– paulplusx
Sep 8 at 11:56
Yup, x=a=b=c,which can't be allowed as I said in an earlier response.But is there a non-calculus way of showing the idea?
– abhibrata ganguly
Sep 8 at 11:58
 |Â
show 1 more comment
1
@prog_SAHIL Please read my statement again. I said if $f(x)$ has $r$ real roots.
– paulplusx
Sep 8 at 11:40
I misread. Deleted.
– prog_SAHIL
Sep 8 at 11:46
How do I proceed?
– abhibrata ganguly
Sep 8 at 11:52
@abhibrataganguly Can you prove $f'(x)$ has no real roots?
– paulplusx
Sep 8 at 11:56
Yup, x=a=b=c,which can't be allowed as I said in an earlier response.But is there a non-calculus way of showing the idea?
– abhibrata ganguly
Sep 8 at 11:58
1
1
@prog_SAHIL Please read my statement again. I said if $f(x)$ has $r$ real roots.
– paulplusx
Sep 8 at 11:40
@prog_SAHIL Please read my statement again. I said if $f(x)$ has $r$ real roots.
– paulplusx
Sep 8 at 11:40
I misread. Deleted.
– prog_SAHIL
Sep 8 at 11:46
I misread. Deleted.
– prog_SAHIL
Sep 8 at 11:46
How do I proceed?
– abhibrata ganguly
Sep 8 at 11:52
How do I proceed?
– abhibrata ganguly
Sep 8 at 11:52
@abhibrataganguly Can you prove $f'(x)$ has no real roots?
– paulplusx
Sep 8 at 11:56
@abhibrataganguly Can you prove $f'(x)$ has no real roots?
– paulplusx
Sep 8 at 11:56
Yup, x=a=b=c,which can't be allowed as I said in an earlier response.But is there a non-calculus way of showing the idea?
– abhibrata ganguly
Sep 8 at 11:58
Yup, x=a=b=c,which can't be allowed as I said in an earlier response.But is there a non-calculus way of showing the idea?
– abhibrata ganguly
Sep 8 at 11:58
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5
It's an increasing function....
– Lord Shark the Unknown
Sep 8 at 11:22
1
Suppose $a$, $b$ and $c$ are real. $(x-a)^3+(x-b)^3+(x-c)^3$ is straightly increasing, so its graph cuts the $x$-axis once only.
– Yuta
Sep 8 at 11:22
Thanks for helping. But can we show that it cuts the x-axis only once?By algebra.I can see that it follows the graph of x^3,But that is not a proof.
– abhibrata ganguly
Sep 8 at 11:27