Ta có : $a+b+c=0$
$\to (a+b+c)^2=0$
$\to a^2+b^2+c^2+2.(ab+bc+ca)=0$
$\to 1+2.(ab+bc+ca)=0$ ( Do $a^2+b^2+c^2=1 $)
$\to ab+bc+ca=-\dfrac{1}{2}$
$\to (ab+bc+ca)^2=\dfrac{1}{4}$
$\to a^2b^2+b^2c^2+c^2a^2+2.(ab^2c+a^2bc+abc^2)=\dfrac{1}{4}$
$\to (ab)^2+(bc)^2+(ca)^2+2abc.(a+b+c)=\dfrac{1}{4}$
$\to (ab)^2+(bc)^2+(ca)^2=\dfrac{1}{4}$ ( Do $a+b+c=0$ )
Ta có : $a^2+b^2+c^2=1$
$\to (a^2+b^2+c^2)^2=1$
$\to a^4+b^4+c^4+2.(a^2b^2+b^2c^2+c^2a^2)=1$
$\to a^4+b^4+c^4 + 2.\dfrac{1}{4}= 1$ ( Do $(ab)^2+(bc)^2+(ca)^2=\dfrac{1}{4}$ )
$\to a^4+b^4+c^4=\dfrac{1}{2}$
Vậy $M=a^4+b^4+c^4=\dfrac{1}{2}$
