Aduceti expresia \( (a+b+c)^4-(a+b)^4-(b+c)^4-(c+a)^4+a^4+b^4+c^4 \) la forma cea mai simpla.
Internet Olympiad, Ariel University, Samaria, Israel
Internet Olympiad Problema 1
Moderators: Filip Chindea, Andrei Velicu, Radu Titiu
- Beniamin Bogosel
- Co-admin
- Posts: 710
- Joined: Fri Mar 07, 2008 12:01 am
- Location: Timisoara sau Sofronea (Arad)
- Contact:
Internet Olympiad Problema 1
Yesterday is history,
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Re: Internet Olympiad Problema 1
Notam \( P(a,b,c)=(a+b+c)^4-(a+b)^4-(b+c)^4-(c+a)^4+a^4+b^4+c^4 \) polinom simetric omogen in a, b, c.
Deoarece P(0,b,c)=0, P(a,0,c)=0 si P(a,b,0)=0, atunci toate monoamele vor contine abc, deci vom avea P(a,b,c)=kabc(a+b+c).
Daca a=b=c=1 se obtine \( 3^4-2^4-2^4-2^4+1+1+1=3k \), deci k=12.
Asadar P(a,b,c)=12abc(a+b+c).
Deoarece P(0,b,c)=0, P(a,0,c)=0 si P(a,b,0)=0, atunci toate monoamele vor contine abc, deci vom avea P(a,b,c)=kabc(a+b+c).
Daca a=b=c=1 se obtine \( 3^4-2^4-2^4-2^4+1+1+1=3k \), deci k=12.
Asadar P(a,b,c)=12abc(a+b+c).