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Cho a+b+c=0 và a^2+b^2+c^2=1

admin - 07/04/2023 82

I tried lớn solve this problem, và I get $a^4+b^4+c^4 = 2(a^2b^2 + 1)$ but I"m not sure if it"s correct

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$egingroup$ it would be better if you can write how did you get that part, then some body can trace errors (if there are any)... $endgroup$
Given $a+b+c = 0$ and $(a^2+b^2+c^2) = 1$, Now $(a^2+b^2+c^2)^2 = 1^2 = 1$

$(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2) = 1.................(1)$

and from $(a+b+c)^2 = 0$,

we get $displaystyle 1+2(ab+bc+ca) = 0Rightarrow (ab+bc+ca) = -frac12$

again squaring both side , we get $displaystyle (ab+bc+ca)^2 = frac14$

$displaystyle (a^2b^2+b^2c^2+c^2a^2)+2abc(a+b+c) = frac14Rightarrow (a^2b^2+b^2c^2+c^2a^2) = frac14$

So put in eqn.... $(1)$ , we get

$displaystyle (a^4+b^4+c^4)+2cdot frac14 = 1Rightarrow (a^4+b^4+c^4) = frac12$

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answered Oct 13, 2013 at 8:38

juantheronjuantheron
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$$a+b+c=0Rightarrow c=-(a+b)$$$$1=a^2+b^2+(a+b)^2=2a^2+2ab+2b^2=2(a^2+ab+b^2)Rightarrow a^2+ab+b^2=frac12$$$$eginalign*a^4+b^4+c^4 &= a^4+b^4+(a^4+4a^3b+6a^2b^2+4ab^3+b^4)\&= 2(a^4+2a^3b+3a^2b^2+2ab^3+b^4)\&= 2(a^2(a^2+ab+b^2)+b^2(a^2+ab+b^2)+ab(a^2+ab+b^2))\&= a^2+ab+b^2=frac12endalign*$$

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answered Oct 13, 2013 at 8:43

Dennis GulkoDennis Gulko
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Given a+b+c=0a+b+c=0 and (a2+b2+c2)=1(a2+b2+c2)=1, Now (a2+b2+c2)2=12=1(a2+b2+c2)2=12=1(a4+b4+c4)+2(a2b2+b2c2+c2a2)=1.................(1)(a4+b4+c4)+2(a2b2+b2c2+c2a2)=1.................(1)and from (a+b+c)2=0(a+b+c)2=0,

we get 1+2(ab+bc+ca)=0⇒(ab+bc+ca)=−121+2(ab+bc+ca)=0⇒(ab+bc+ca)=−12again squaring both side , we get (ab+bc+ca)2=14(ab+bc+ca)2=14(a2b2+b2c2+c2a2)+2abc(a+b+c)=14⇒(a2b2+b2c2+c2a2)=14(a2b2+b2c2+c2a2)+2abc(a+b+c)=14⇒(a2b2+b2c2+c2a2)=14So put in eqn.... (1)(1) , we get

(a4+b4+c4)+2⋅14=1⇒(a4+b4+c4)=12

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answered Apr 18, năm 2016 at 14:15