$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?

Question

$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?

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

Answer 1

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 $\displaystyle 1+2(ab+bc+ca) = 0\Rightarrow (ab+bc+ca) = -\frac{1}{2}$

again squaring both side , we get $\displaystyle (ab+bc+ca)^2 = \frac{1}{4}$

$\displaystyle (a^2b^2+b^2c^2+c^2a^2)+2abc(a+b+c) = \frac{1}{4}\Rightarrow (a^2b^2+b^2c^2+c^2a^2) = \frac{1}{4}$

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

$\displaystyle (a^4+b^4+c^4)+2\cdot \frac{1}{4} = 1\Rightarrow (a^4+b^4+c^4) = \frac{1}{2}$

Answer 2

$$a+b+c=0\Rightarrow c=-(a+b)$$ $$1=a^2+b^2+(a+b)^2=2a^2+2ab+2b^2=2(a^2+ab+b^2)\Rightarrow a^2+ab+b^2=\frac12$$ $$\begin{align*} a^4+b^4+c^4 &= a^4+b^4+(a^4+4a^3b+6a^2b^2+4ab^3+b^4)\\ &= 2(a^4+2a^3b+3a^2b^2+2ab^3+b^4)\\ &= 2(a^2(a^2+ab+b^2)+b^2(a^2+ab+b^2)+ab(a^2+ab+b^2))\\ &= a^2+ab+b^2=\frac12 \end{align*}$$

Answer 3

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)=−12 again 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)=14 So put in eqn.... (1)(1) , we get

(a4+b4+c4)+2⋅14=1⇒(a4+b4+c4)=12