Can a^3+b^3+c^3 Equal d^3+e^3 Given Specific Conditions?

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SUMMARY

The discussion centers on the mathematical problem of determining whether the expression \(a^3 + b^3 + c^3\) can equal \(d^3 + e^3\) under the constraints \(a^2 + b^2 + c^2 = d^2 + e^2\) and \(a^4 + b^4 + c^4 = d^4 + e^4\). Participants explore various scenarios, including when all variables are greater than 1, equal to 1, or between 0 and 1. Key insights reveal that if \(a = b = c = 1\), contradictions arise, indicating that this case is not feasible. The discussion suggests further investigation into the implications of limiting the domain of the variables.

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Problem:

Let $a$, $b$, $c$, $d$ and $e$ be strictly positive real numbers such that:

$$a^2+b^2+c^2=d^2+e^2$$

$$a^4+b^4+c^4=d^4+e^4$$

Compare $$a^3+b^3+c^3$$ with $$d^3+e^3$$.

I have been exhausting all kinds of algebraic tricks, but I still don't get anywhere near to cracking it, and it is so frustrating not knowing how to solve it...

Any help and/or suggestions toward how to solve this problem is much appreciated.
 
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Have you considered limiting the Domain?

1) What happens when all of a, b, c, d, e are greater than 1?
2) Everything is 1
3) All are between 0 and 1.
 
First, I see some restrictions on $a, b$ and $c$. If you eliminate say $e$ from the pair of equations you get

$d^4 - (a^2+b^2+c^2)d^2 + a^2b^2 + a^2c^2 + b^2c^2 = 0$

Since $d$ is real this means that

$(a^2+b^2+c^2)^2-4(a^2b^2 + a^2c^2 + b^2c^2) \ge 0$

or

$a^4+b^4+c^4- 2 a^2b^2- 2 a^2c^2 -2 b^2c^2 \ge 0$

Also note that if $a=b=c = 1$ then we have a contradiction so this can't happen.
 
Last edited:
tkhunny said:
Have you considered limiting the Domain?

1) What happens when all of a, b, c, d, e are greater than 1?
2) Everything is 1
3) All are between 0 and 1.

Thanks for the hint, tkhunny...and I think we can rule out the possibility when all of them are 1 because

$$1^2+1^2+1^2 \ne 1^2+1^2$$ and $$1^4+1^4+1^4 \ne 1^4+1^4$$

But for the case where all of the variables are between 0 and 1, do you mean to suggest that we could let, for example a=cosA as another way to approach the problem?

Jester said:
First, I see some restrictions on $a, b$ and $c$. If you eliminate say $e$ from the pair of equations you get

$d^4 - (a^2+b^2+c^2)d^2 + a^2b^2 + a^2c^2 + b^2c^2 = 0$

Since $d$ is real this means that

$(a^2+b^2+c^2)^2-4(a^2b^2 + a^2c^2 + b^2c^2) \ge 0$

or

$a^4+b^4+c^4- 2 a^2b^2- 2 a^2c^2 -2 b^2c^2 \ge 0$

Also note that if $a=b=c = 1$ then we have a contradiction so this can't happen.

Thanks Jester for showing me some insight that I would not have thought of it myself...but, what should I do after that? I am so confused...
 

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