1no prompt replies
10 Answers
1its just given
a+b+c=0
take them to be on the real line
i guess something
using 'complex' has to be done
1Well , I found out the simplest { and the most boringg as well } solution -
a 5 + b 5 + c 5
= a 5 + b 5 - { a + b } 5
= - 5 a 4 b - 10 a 3 b 2 - 10 a 2 b 3 - 5 a b 4
= - 5 a b { a 3 + b 3 } - 10 a 2 b 2 { a + b }
= - 5 a b { a + b } { a 2 - a b + b } - 10 a 2 b 2 { a + b }
= - 5 a b { a + b } [ a 2 + b 2 - a b + 2 a b ]
= - 5 a b { a + b } { a 2 + a b + b 2 }
a 3 + b 3 + c 3 = - 3 a b { a + b }
a 2 + b 2 + c 2 = 2 { a 2 + b 2 + a b }
Put the VALUES , and you get the answer .
341Let a,b,c be roots of the cubic
x^3-px^2+qx-r = 0
Then p = 0 and hence the cubic is x^3+qx-r = 0. Remember that q = \sum ab; r = abc
Since a is a root of the cubic we have a^3+qa-r = 0 \Rightarrow a^5+qa^3-ra^2 = 0
\Rightarrow \sum a^5+q\sum a^3+-r\sum a^2 = 0 \Rightarrow \sum a^5 = -q\sum a^3+r\sum a^2
Since a+b+c = 0, q = -\frac{\sum a^2}{2}; r = \frac{\sum a^3}{3}
Which gives \sum a^5 = \frac{5\sum a^2\sum a^3}{6} or written in the more symmetrical form
\frac{\sum a^5}{5} = \frac{\sum a^2}{2} \ \times \ \frac{\sum a^3}{3}
1thanks prophet sir
line 5
elaborated more
(a+b+c)^2 = \sum{a^2}+\underline{2\sum{ab}} _p
\texttt{if a+b+c=0 , we have } \\ a^3+b^3+c^3 =3abc
any solution through some inequality ? is it possible ?
another solution that was given in book was using log deries expansion
62yup i guess it is a problem that i had seen probably in a very less known book for Olympiads by pitambar publications! (I used that book when i was studying for olympiads)
62@Gallardo.. Your second proof is not correct. In fact far from it if i have understood it correctly... I guess you have used an example to give a proof!
62another method which i always suggest is to reduce the number of variables...
Divide by a^5 to get... x+y+1=0
\frac{(1+x^5+y^5)}{5}=\frac{(1+x^3+y^3)}{3}\frac{(1+x^2+y^2)}{2}
Now it becomes slightly simpler to work \frac{(1+(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4))}{5}=\frac{(1+(x+y)(x^2+y^2-xy))}{3}\frac{(1+x^2+y^2+2xy-2xy)}{2}
Now it becomes slightly simpler to work \frac{(1-(x^4-x^3y+x^2y^2-xy^3+y^4))}{5}=\frac{(1-(x^2+y^2-xy))}{3}\frac{(1+1-2xy)}{2}
\\\frac{(1-(x^4-x^3y+x^2y^2-xy^3+y^4))}{5}=\frac{(1-(x^2+y^2-xy))}{3}\frac{(1+1-2xy)}{2} \\or \frac{(1-((x^2+y^2)^2-xy(x^2+y^2)-x^2y^2)}{5}=\frac{(1-((x+y)^2-3xy))}{3}\frac{(1+1-2xy)}{2} \\or \frac{(1-(x^2+y^2)(x^2+y^2-xy)-x^2y^2)}{5}=\frac{(1-(1-3xy))}{3}(1-xy) \\or \frac{(1-(1-2xy)(1-3xy)-x^2y^2)}{5}=xy(1-xy)
Which is always true...
Now the proof becomes slightly (only) better than gallardo's :)