Question

If $a+b+c=1,ab+bc+ca=2$ and $abc=3$, then the value of $a^4+b^4+c^4$ is equal to

Answer 1

$\because a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)$
$\therefore a^2+b^2+c^2=1-4=-3$
and $a^2{b}^2+b^2{c}^2+c^2{a}^2=(ab+bc+ca{)}^2-2abc(a+b+c)$
$a^2{b}^2+b^2{c}^2+c^2{a}^2=4-6=-2$
Now, $a^4+b^4+c^4=(a^2+b^2+c^2{)}^2-2(a^2{b}^2+b^2{c}^2+c^2{a}^2)$
$\Rightarrow a^4+b^4+c^4=9+4$
$\therefore a^4+b^4+c^4=13$