29. If $a, b, c$ are positive real numbers, solve the equation
$$
\begin{array}{l}
\frac{x^{8}+a^{8}}{(x+a)^{3}}+\frac{x^{3}+b^{3}}{(x+b)^{3}}+\frac{x^{3}+c^{3}}{(x+c)^{8}} \\
=\frac{3}{2}-\frac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)} .
\end{array}
$$
29. Notice that
$$
\left(x^{3}+a^{3}\right)=(x+a)\left((x+a)^{2}-3 a x\right),
$$

Therefore, the given equation can be rewritten as:
$$
\begin{array}{l}
\frac{a x}{(x+a)^{2}}+\frac{b x}{(x+b)^{2}}+\frac{c x}{(x+c)^{2}} \\
=\frac{x^{8}+(a b+b c+c a) x}{(x+a)(x+b)(x+c)} .
\end{array}
$$

Thus, one root of the equation is $x=0$, and moreover, we have
$$
\begin{array}{l}
a(x+b)^{2}(x+c)^{2}+b(x+c)^{2} \\
-(x+a)^{2}+c(x+a)^{2}(x+b)^{2} \\
=\left(x^{2}+a b+b c+c a\right)(x+a) \\
\cdot(x+b)(x+c),
\end{array}
$$

which simplifies to
$$
x\left[x^{4}-2(a b+b c+c a) x^{2}-8 a b c x+a^{2} b^{2}\right.
$$
$$
\left.+b^{2} c^{2}+c^{2} a^{2}-2 a^{2} b c-2 b^{2} c a-2 c^{2} a b\right]=0 \text {. }
$$

Thus, $x=0$ is a double root of the equation. Setting the second factor to 0 and simplifying, we get
$$
\left(x^{2}-a b-b c+c a\right)^{2}-(2 \sqrt{c a}(x+b))^{2}=0 \text {. }
$$

Therefore,
$$
\begin{array}{l}
x^{2} \pm 2 \sqrt{c a} x+c a-a b \pm 2 b \sqrt{c a}-b c \\
=(x \pm \sqrt{c a})^{2}-(\sqrt{a b} \mp \sqrt{b c})^{2} .
\end{array}
$$

Thus, the remaining four roots are
$$
\begin{array}{l}
x=\sqrt{a b}+\sqrt{b \bar{c}}+\sqrt{c a}, \\
x=\sqrt{a \bar{b}}-\sqrt{b \overline{b c}}-\sqrt{c a} \text {, } \\
x=-\sqrt{a b}+\sqrt{b} \bar{c}-\sqrt{c a}, \\
x=-\sqrt{a b}-\sqrt{b c}+\sqrt{c a} \text {. } \\
\end{array}
$$

If $a \neq b \neq c$, all six roots are valid. If $a=b=c$, then there are only three roots, namely the double root $x=0$ and $x=\sqrt{a b}+\sqrt{\bar{b} c}+\sqrt{c a}=3 a$. The other three roots all reduce to $x=-a$, which must be discarded (since the original equation has terms involving $x+a$).
If $a=b \neq c$, then in addition to the double root $x=0$, we have
$$
\begin{array}{l}
x=\sqrt{a b}+\sqrt{b c}+\sqrt{c a}=a+2 \sqrt{c a}, \\
x=\sqrt{a} \bar{b}-\sqrt{b \bar{b}}-\sqrt{c a}=a-2 \sqrt{c a} .
\end{array}
$$

The other two roots all reduce to $x=-a$, which must also be discarded. The cases $b=c \neq a$ and $c=a \neq b$ are similar, each yielding four roots.