MHB Can this system of equations be solved in real numbers?

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The system of equations given is $a(b+c-a^3)=b(c+a-b^3)=c(a+b-c^3)=1$. A solution approach involves simplifying the equations to $ab + ac - a^4 = 1$, $ab + bc - b^4 = 1$, and $ac + bc - c^4 = 1$. The discussion highlights that the solution was found through inspection rather than extensive calculation. Participants are encouraged to share their methods for finding solutions. The focus remains on solving the equations in real numbers.
anemone
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Solve in real numbers the system below:

$a(b+c-a^3)=b(c+a-b^3)=c(a+b-c^3)=1$
 
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a=1, b=1, c=1
a=-1, b=-1, c=-1
 
Wilmer said:
a=1, b=1, c=1
a=-1, b=-1, c=-1

Would you mind sharing how you found the solution? :D
 
Lazily, by inspection:
ab + ac - a^4 = 1
ab + bc - b^4 = 1
ac + bc - c^4 = 1
 
Wilmer said:
a=1, b=1, c=1
a=-1, b=-1, c=-1

Hi Wilmer,

Your answer (without the working, hehehe...) is correct, but the question remains on how we are going to prove those are the only solutions.(Nod)
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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