Question

If $a+b+c=0$ & $a^2+b^2+c^2=1$, then the value of $a^4+b^4+c^4$ is

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{2}$

Given $a+b+c=0$ and $a^2+b^2+c^2=1$
$a+b+c=0$
Squaring on both sides,
$a^2+b^2+c^2+2(ab+bc+ca)=0$
$\Rightarrow ab+bc+ca=\frac{-1}{2}$
Squaring on both sides,
$a^2{b}^2+b^2{c}^2+c^2{a}^2+2abc(a+b+c)=\frac{1}{4}$
$\Rightarrow a^2{b}^2+b^2{c}^2+c^2{a}^2=\frac{1}{4}$
Now, $a^4+b^4+c^4=(a^2+b^2+c^2{)}^2-2(a^2{b}^2+b^2{c}^2+c^2{a}^2)$
$\therefore a^4+b^4+c^4=1-2\left(\frac{1}{4}\right)=\frac{1}{2}$
Hence, option C.