Giải thích các bước giải:
Ta có :
$a+b+c=0\to a=-(b+c)\to a^2=(b+c)^2$
Tương tự $b^2=(a+c)^2,c^2=(a+b)^2$
$\to xa^2+yb^2+zc^2=x(b+c)^2+y(a+c)^2+z(a+b)^2$
$\to xa^2+yb^2+zc^2=x(b^2+2bc+c^2)+y(a^2+2ac+c^2)+z(a^2+2ab+b^2)$
$\to xa^2+yb^2+zc^2=a^2(y+z)+b^2(x+z)+c^2(x+y)+2(xbc+yac+zab)$
$\to xa^2+yb^2+zc^2=a^2(y+z)+b^2(x+z)+c^2(x+y)+2abc(\dfrac xa+\dfrac yb+\dfrac zc)$
$\to xa^2+yb^2+zc^2=a^2(y+z)+b^2(x+z)+c^2(x+y)+2abc.0$
$\to xa^2+yb^2+zc^2=a^2(-x)+b^2(-y)+c^2(-z)$
$\to xa^2+yb^2+zc^2=-(xa^2+yb^2+zc^2)$
$\to 2(xa^2+yb^2+zc^2)=0$
$\to xa^2+yb^2+zc^2=0$
