7. Given non-zero real numbers $a, b, c$ satisfy:
$$
a+b+c=0, a^{4}+b^{4}+c^{4}=128 \text{. }
$$

Find all possible values of $a b+b c+c a$.


7. Notice that,
$$
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \text {, }
$$

and $a+b+c=0$.
$$
\begin{array}{l} 
\text { Then } a^{2}+b^{2}+c^{2}=-2(a b+b c+c a) . \\
\text { By }(a+b+c)\left(a^{3}+b^{3}+c^{3}\right) \\
= a^{4}+b^{4}+c^{4}+a b\left(a^{2}+b^{2}\right)+ \\
b c\left(b^{2}+c^{2}\right)+c a\left(c^{2}+a^{2}\right) \\
= a^{4}+b^{4}+c^{4}+(a b+b c+c a)\left(a^{2}+b^{2}+c^{2}\right)- \\
a b c(a+b+c),
\end{array}
$$

Substituting $a+b+c=0$ into the above equation, we get
$$
\begin{array}{l}
128=a^{4}+b^{4}+c^{4}=2(a b+b c+c a)^{2} \\
\Rightarrow a b+b c+c a= \pm 8 .
\end{array}
$$

Since $a b+b c+c a=-\frac{a^{2}+b^{2}+c^{2}}{2} \leqslant 0$, we have
$$
a b+b c+c a=-8 .
$$