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

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The discussion centers on the mathematical problem of determining whether the sum of cubes, a^3 + b^3 + c^3, can equal d^3 + e^3 under specific conditions where a, b, c, d, and e are positive real numbers satisfying 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 suggest that if all variables equal 1, contradictions arise, indicating this case is impossible. The discussion also highlights the importance of algebraic manipulation and potential domain restrictions to further analyze the problem. The complexity of the problem leaves participants seeking additional strategies to reach a solution.
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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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Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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