17-128 Given that $a, b, c$ are the lengths of the three sides of a triangle, then the value of $a^{4}+b^{4}+c^{4}-$ $2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2}$ is
(A) always positive.
(B) always negative.
(C) can be positive or negative.
(D) non-negative.
(China Shanxi Province Taiyuan City Junior High School Mathematics Competition, 1991)
[Solution 1]
$$
\begin{aligned}
\text { [Solution 1] } & a^{4}+b^{4}+c^{4}-2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2} \\
= & a^{4}+\left(b^{2}-c^{2}\right)^{2}-\left[a^{2}(b+c)^{2}+a^{2}(b-c)^{2}\right] \\
= & a^{4}-\left[a^{2}(b+c)^{2}+a^{2}(b-c)^{2}\right]+(b+c)^{2}(b-c)^{2} \\
= & {\left[a^{2}-(b+c)^{2}\right]\left[a^{2}-(b-c)^{2}\right] } \\
= & (a-b-c)(a+b+c)(a-b+c)(a+b-c), \\
\because \quad & a+b>c, \quad a+c>b, \quad b+c>a,
\end{aligned}
$$
$\therefore$. The above expression is always less than 0.
Therefore, the answer is $(B)$.
[Solution 2] By the square of a trinomial formula, we have
$$
\begin{aligned}
\text { Original expression } & =\left(a^{2}+b^{2}-c^{2}\right)^{2}-4 a^{2} b^{2} \\
& =(a+b+c)(a+b-c)(a-b+c)(a-b-c),
\end{aligned}
$$

The remaining steps are the same as above.
Therefore, the answer is $(B)$.