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- Feb 14, 2012

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I have solved the problem as stated below but I don't know if it's an unique solution and even if it is, I have no idea how to prove that would be the case.

Can anyone show me how to approach the problem correctly?

For the equation $x^5-12x^4+ax^3+bx^2+cx-64=0$, all of its roots are positive real numbers. Evaluate the sum of $a+b+c$.

Attempt:

It's quite obvious from the values of the product of all 5 roots and the sum of them reveal that the equation $x^5-12x^4+ax^3+bx^2+cx-64=0$ has roots of 2 and 4, of which 2 is the repeated root of multiplicity 4, since

$2+2+2+2+4=12$ and $2^4(4)=64$,

Then there are many ways to find the values for $a, b, c$ and at last, after the values of $a, b, c$ are known, we can conclude that $a+b+c=56+144-128=72$.