$a+b+c=0;a^2+b^2+c^2=1$ 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

$a,b,c$ are the three roots of a cubic equation $$x^3+\alpha x^2+\beta x+\gamma=0$$ where $$\alpha=-(a+b+c)=0\quad\text{and}\quad\beta=ab+bc+ca=\frac{(a+b+c)^2-(a^2+b^2+c^2)}2=-\frac12.$$ Each of them satisfies $$x^4=-(\alpha x^2+\beta x+\gamma)x=\frac{x^2}2-\gamma x$$ hence $$a^4+b^4+c^4=\frac{a^2+b^2+c^2}2-\gamma(a+b+c)=\frac12.$$

Answer 4

Given $a + b + c = 0$ and $(a^2 + b^2 + c^2) = 1$,

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

Expanding, we have: $$ (a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2) = 1. \tag{1} $$

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

we get: $$ 1 + 2(ab + bc + ca) = 0 \implies ab + bc + ca = -\frac{1}{2}. $$

Squaring both sides, we obtain: $$ (ab + bc + ca)^2 = \left( -\frac{1}{2} \right)^2 = \frac{1}{4}. $$

Thus, $$ a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c) = \frac{1}{4}. $$

Given $a + b + c = 0$, the term $2abc(a + b + c) = 0$, so: $$ a^2b^2 + b^2c^2 + c^2a^2 = \frac{1}{4}. $$

Substituting this result into equation $(1)$, we have: $$ a^4 + b^4 + c^4 + 2 \cdot \frac{1}{4} = 1. $$

Simplifying, we get: $$ a^4 + b^4 + c^4 + \frac{1}{2} = 1. $$

Therefore, $$ a^4 + b^4 + c^4 = 1 - \frac{1}{2} = \frac{1}{2}. $$

Thus, the final result is: $$ \boxed{a^4 + b^4 + c^4 = \frac{1}{2}}. $$