So x=0 is the real root of f(x)=0.

Let X 1=0 and f(x2)=0 because x2 is the root of the function.

Then: f(X2)=-f(-x2)=0 is odd function property, so there is f(-x2)=0, that is, -x2 is also a root, and similarly -x3 is also a root.

Because the function f has only three real roots.

1: If -x2=X3，X 1+X2+X3=0。

2,: If -x2=x 1, then x2=x 1, →x3 must be equal to x 1, x2, otherwise the equation will have four roots, namely x 1, x2, x3, -x3,

X 1+x2+x3=0 is deduced.