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[obm-l] Re: [obm-l] Questão



      Se a + b + c = 0 e   a^2 + b^2 + c^n = 1  , então          
          a^4 + b^4 + c^4 = ?
 
(a + b + c)^2 = 0 = a^2 + b^2 + c^2 + 2 (ab + bc + ac) 
 
0 = a^2 + b^2 + c^2 + 2 (ab + bc + ac) 
a^2 + b^2 + c^n = 1
 
0 = 1+ 2 (ab + bc + ac) 
(ab + bc + ac) = -1/2
(ab + bc + ac)^2 = 1/4
( (ab)^2 + (bc)^2 + (ac)^2 + 2 ( abcb + aabc + abcc )) = 1/4
( (ab)^2 + (bc)^2 + (ac)^2 + 2abc (a + b + c) ) = 1/4
( (ab)^2 + (bc)^2 + (ac)^2 + 2abc (0) ) = 1/4
( (ab)^2 + (bc)^2 + (ac)^2 ) = 1/4
 
 
(a^2 + b^2 + c^2)^2=1^2
a^4 + b^4 + c^4 + 2((ab)^2 + (bc)^2 + (ac)^2)=1
a^4 + b^4 + c^4 + 2(1/4)=1
a^4 + b^4 + c^4 + 1/2=1
a^4 + b^4 + c^4 =1/2
 
 
Longo mas funcional... rsrs espero q esteja certo. Abraços
 
MuriloRFL
 
 
----- Original Message -----
Sent: Monday, December 12, 2005 1:28 PM
Subject: [obm-l] Questão

      Se a + b + c = 0 e   a^2 + b^2 + c^n = 1  , então          
          a^4 + b^4 + c^4 = ?