The problem involves proving the equation 6(a^5 + b^5 + c^5) = 5(a^3 + b^3 + c^3)(a^2 + b^2 + c^2) under the condition that a + b + c = 0. The subject area pertains to algebraic identities and polynomial equations.
The discussion is ongoing, with participants exploring different algebraic approaches to prove the equation. Some have provided guidance on how to manipulate the equation, while others question the validity of the original statement based on their numerical examples.
There is a noted emphasis on the condition a + b + c = 0, which is central to the problem. Some participants express confusion regarding the relationship between the two sides of the equation when specific values are substituted.
Please read the question carefully. It says that the above holds only if a+b+c=0 and that is what I want to prove.Vacrin said:Im sorry to say, but none of these equations work, they are not equal so they have no solutions.
I did a quick test by putting numbers in for a b and c. Being 2,3,4 the first part 6(a^5+b^5+c^5) = 5(a^3+b^3+c^3)(a^2+b^2+c^2) using the numbers i offered earlier. It comes to 7794=14355.
The second equation comes to 99=72
I do not want to solve for solutions . ( By the way the solutions are all a , b and c that sum to 0) . But I want to prove that equation when a+b+c=0.Vacrin said:Well, that equation has quite a few solutions, wouldn't you need to solve for them all? Or does mater in the equations given below?
agoogler said:I do not want to solve for solutions . ( By the way the solutions are all a , b and c that sum to 0) . But I want to prove that equation when a+b+c=0.
agoogler said:Please read the question carefully. It says that the above holds only if a+b+c=0 and that is what I want to prove.
Thanks a lot !Ray Vickson said:Put c = -a-b on both sides and expand both sides.